Question 1

An unbiased six faced die is tossed once. What is the probability that the number appearing on the upper face is a multiple of 3?

Accepted Answer

Introduction / Context: This question is a simple probability problem involving the outcomes of rolling a single fair die. We are asked for the probability that the outcome is a multiple of 3. Problems like this provide practice in identifying favorable outcomes in a small finite sample space and applying the basic probability formula for equally likely outcomes. Given Data / Assumptions: The die is fair and has six faces. The faces are numbered 1, 2, 3, 4, 5 and 6. All six outcomes are equally likely. We are looking for outcomes that are multiples of 3. Concept / Approach: We determine which outcomes from the set {1, 2, 3, 4, 5, 6} are multiples of 3. The probability of the event is then the count of these favorable outcomes divided by the total number of possible outcomes, which is 6. This is a direct application of the classical probability definition for equally likely outcomes. Step-by-Step Solution: Step 1: List all possible outcomes from rolling the die: 1, 2, 3, 4, 5 and 6.Step 2: Identify which of these numbers are multiples of 3. Multiples of 3 in this range are 3 and 6.Step 3: Count the favorable outcomes. There are 2 favorable outcomes, namely 3 and 6.Step 4: Count the total number of outcomes, which is 6.Step 5: Probability of getting a multiple of 3 is favorable outcomes divided by total outcomes = 2 / 6.Step 6: Simplify 2 / 6 by dividing numerator and denominator by 2 to obtain 1 / 3. Verification / Alternative check: As a quick check, we can also look at the complement event, which is the event that the outcome is not a multiple of 3. Non multiples of 3 in this set are 1, 2, 4 and 5, which are 4 out of 6 outcomes, giving probability 4 / 6 = 2 / 3. Adding the probabilities of multiples and non multiples of 3, that is 1 / 3 + 2 / 3, yields 1, confirming that our classification and probability calculations are complete and consistent. Why Other Options Are Wrong: 1/2 would imply that half of the die faces are multiples of 3, which would require 3 favorable outcomes, but there are only 2. 3/4 and 3/2 exceed 1/2 and 1 respectively, and therefore cannot represent this event. 2/3 is the probability of obtaining a non multiple of 3, not the probability of obtaining a multiple of 3, so it is the complement of the desired probability. Common Pitfalls: Some students may mistakenly count 0 as an outcome or misidentify the multiples of 3. Others might try to overcomplicate the problem by thinking of divisibility patterns without simply listing the small set of outcomes. In basic dice problems, it is safest to explicitly write down the faces and count the ones that satisfy the required property. Final Answer: The probability that the number appearing on the die is a multiple of 3 is 1/3.

Question 2

A bag contains 40 balls in total. There are 18 red balls and the rest are green or blue balls. Two balls are picked from the bag one after another without replacement. The probability that the first ball is red and the second ball is green is 3/26. …

Accepted Answer

Introduction / Context: This problem combines probability with algebraic reasoning. We know the composition of red balls in a bag and that the remaining balls are green or blue. We are given the probability that the first ball drawn is red and the second ball is green, both drawn without replacement. Using this probability, we must determine how many blue balls are in the bag. Given Data / Assumptions: Total number of balls in the bag = 40. Number of red balls = 18. The remaining 22 balls are either green or blue. Two balls are drawn in sequence without replacement. The probability that the first ball is red and the second is green is 3/26. We assume all balls of the same color are indistinguishable and that each ball is equally likely to be drawn. Concept / Approach: Let the number of green balls be G and the number of blue balls be B. We know that G + B = 22. The probability that the first ball is red and the second is green can be expressed in terms of G. Specifically, it is (18/40) multiplied by (G/39). We equate this product to 3/26 and solve for G. Once G is known, we compute B as 22 minus G to find the number of blue balls. Step-by-Step Solution: Step 1: Let G be the number of green balls and B be the number of blue balls, so G + B = 22.Step 2: The probability that the first ball is red is 18 / 40.Step 3: Given that the first ball drawn is red, there are now 39 balls left in the bag, of which G are green.Step 4: The conditional probability that the second ball is green is therefore G / 39.Step 5: The overall probability of first red and then green is (18 / 40) * (G / 39), and this is given to be 3 / 26.Step 6: Set up the equation (18 / 40) * (G / 39) = 3 / 26 and solve for G. Simplifying, we obtain G = 10, so B = 22 − 10 = 12 blue balls. Verification / Alternative check: We can verify by substituting G = 10 back into the probability expression. The probability of first drawing red and then green is (18/40) * (10/39). Multiply numerators to get 180 and denominators to get 1560. Simplifying 180 / 1560 by dividing both numerator and denominator by 60 gives 3 / 26, which matches the given probability. This confirms that the counts 10 green and 12 blue are consistent with the conditions. Why Other Options Are Wrong: 16 blue balls would imply only 6 green balls, which would lead to a smaller probability than 3/26. 10 blue balls would leave 12 green balls, which yields (18/40) * (12/39) = 216/1560, simplifying to 9/65, not 3/26. 14 or 8 blue balls would similarly give probabilities that do not match 3/26 when used in the same calculation. Therefore, only 12 blue balls produce the correct probability. Common Pitfalls: Students may incorrectly set up the equation by reversing the order of colors or by using 40 in both denominators, forgetting the reduction to 39 after the first draw. Another common error is to attempt a guess and check method with the options rather than solving the equation algebraically, which can be time consuming and error prone. Carefully defining variables and writing the probability in terms of those variables is the safest approach. Final Answer: The number of blue balls in the bag is 12.

Question 3

There are four different hotels in a town. Three different men check into hotels in a day, each choosing a hotel independently and at random. What is the probability that each man checks into a different hotel, so that no two men are in the same hot…

Accepted Answer

Introduction / Context: This question involves discrete probability with independent choices among several options. Each of three men chooses one of four hotels independently. We are asked for the probability that all three choose different hotels, meaning that no hotel is chosen by more than one man. This problem illustrates counting outcomes when order matters and then forming a probability ratio. Given Data / Assumptions: There are exactly 4 distinct hotels in the town. There are 3 distinct men making choices. Each man chooses exactly one hotel. Each man chooses a hotel independently and each hotel is equally likely for each man. We want the probability that all three men end up in three different hotels. Concept / Approach: We treat each man's choice as an independent selection from 4 options, so the total number of ordered outcomes is 4^3. To find favorable outcomes, we need to count the number of ordered triples in which all three chosen hotels are distinct. This is equivalent to counting permutations of 4 hotels taken 3 at a time, because each man must be assigned a different hotel and order matters. Step-by-Step Solution: Step 1: For each man, there are 4 choices of hotel, so the total number of ordered outcomes is 4 * 4 * 4 = 4^3 = 64.Step 2: To have all different hotels, the first man can choose any of the 4 hotels.Step 3: Once the first man has chosen, the second man must choose a different hotel, leaving 3 choices.Step 4: The third man must choose a hotel different from both previous choices, leaving 2 choices.Step 5: The number of favorable outcomes is therefore 4 * 3 * 2 = 24.Step 6: The required probability is favorable outcomes divided by total outcomes, giving 24 / 64, which simplifies to 3 / 8 after dividing numerator and denominator by 8. Verification / Alternative check: We can cross check by computing the probability directly using conditional reasoning. The first man picks any hotel, which has probability 1. The second man must pick a different hotel, which has probability 3/4. The third man must pick a different hotel from the two already chosen, which has probability 2/4 = 1/2. Multiplying these gives 1 * (3/4) * (1/2) = 3/8, which matches the result obtained from counting permutations. This independently confirms the answer. Why Other Options Are Wrong: 1/2 would result from incorrectly assuming that half of the outcomes are favorable, which is not supported by the detailed counts. 3/4 would require 48 favorable outcomes, but we only have 24. 4/7 and 9/16 have denominators that do not match 64 when scaled and do not correspond to any simple product of 4 and 3 in this context. Only 3/8 arises naturally from correct counting and conditional probability reasoning. Common Pitfalls: Students may mistakenly treat the men as indistinguishable, leading to undercounting. Others may forget that order matters when counting total outcomes, or they might simply think about combinations of hotels without accounting for which man goes to which hotel. Working step by step with conditional probabilities or using permutations of hotels taken three at a time helps avoid these errors. Final Answer: The probability that each man checks into a different hotel is 3/8.

Question 4

In a race, the odds in favour of cars P, Q, R and S winning are given respectively as 1:3, 1:4, 1:5 and 1:6. Assuming these odds correctly represent their chances against all other competitors, what is the probability that one of these four cars (P,…

Accepted Answer

Introduction / Context: This problem involves interpreting odds in favour of different cars winning a race and converting those odds into probabilities. Four cars P, Q, R and S each have odds in favour of winning given against all other competitors. We must find the probability that one of these four cars wins the race, which requires converting each set of odds into a probability and then adding them together. Given Data / Assumptions: Odds in favour of car P winning are 1:3. Odds in favour of car Q winning are 1:4. Odds in favour of car R winning are 1:5. Odds in favour of car S winning are 1:6. Odds in favour a:b mean probability of event = a / (a + b). We assume the events of each car winning are mutually exclusive and the odds are consistent with an underlying probability model. Concept / Approach: The main idea is to convert odds in favour of each car into probabilities. For odds a:b in favour, the probability that the event occurs is a / (a + b). We do this for each car, obtaining P(P wins), P(Q wins), P(R wins) and P(S wins). Because only one car can win the race, and the events are mutually exclusive, the probability that any one of the four wins is the sum of these four probabilities. Step-by-Step Solution: Step 1: For odds 1:3 in favour of P, the probability P(P wins) = 1 / (1 + 3) = 1 / 4.Step 2: For odds 1:4 in favour of Q, the probability P(Q wins) = 1 / (1 + 4) = 1 / 5.Step 3: For odds 1:5 in favour of R, the probability P(R wins) = 1 / (1 + 5) = 1 / 6.Step 4: For odds 1:6 in favour of S, the probability P(S wins) = 1 / (1 + 6) = 1 / 7.Step 5: The probability that one of these four cars wins is the sum: 1/4 + 1/5 + 1/6 + 1/7.Step 6: Compute this sum using a common denominator. The least common multiple of 4, 5, 6 and 7 is 420. The sum becomes (105 + 84 + 70 + 60) / 420 = 319 / 420. Verification / Alternative check: We can verify the arithmetic by checking each term carefully. Converting to denominator 420 gives 1/4 = 105/420, 1/5 = 84/420, 1/6 = 70/420 and 1/7 = 60/420. Adding these numerators gives 105 + 84 + 70 + 60 = 319, so the total probability is 319/420. This is less than 1, as it must be, which is sensible since the remaining probability accounts for all other competitors in the race who are not among P, Q, R and S. Why Other Options Are Wrong: 27/111 is an oddly simplified fraction that does not result from the natural denominators involved in the odds. 114/121 is greater than 1, which is impossible for a probability. 231/420 would correspond to the sum of only some of the component probabilities and is therefore too small. 1/4 is simply the probability for car P alone and does not include the contributions from Q, R and S. Common Pitfalls: Students frequently misinterpret odds as probabilities, writing 1:3 as 1/3 when the actual probability is 1/4. It is also easy to forget to sum all four probabilities or to ignore that the events are mutually exclusive. Another common mistake is to simplify incorrectly when using a common denominator. Keeping track of each term and using a systematic method to add fractions avoids these errors. Final Answer: The probability that one of the cars P, Q, R or S wins the race is 319/420.

Question 5

Fourteen distinct persons are seated randomly around a circular table. What is the probability that three particular persons always sit together as neighbours, in any order, while the remaining persons occupy the other seats?

Accepted Answer

Introduction / Context: This question combines circular permutations with basic probability. We are given 14 distinct persons seated around a round table and we want the probability that three particular persons always sit together as neighbours. In circular arrangements, rotating everyone together does not create a new arrangement, so we must use the idea of circular permutations. The key concept is to treat the three special persons as a single block and then count how many such favourable arrangements exist compared with the total possible arrangements of all persons. Given Data / Assumptions: There are 14 distinct persons in total. Three of these persons are special and must sit together as neighbours. The table is circular, so arrangements that differ only by rotation are considered the same. All circular seating arrangements are equally likely. Concept / Approach: For n distinct persons seated around a circular table, the number of possible arrangements is (n - 1)!. This is because we can fix one person as a reference to break circular symmetry and then arrange the remaining persons linearly. For the favourable cases, we treat the three special persons as one combined block and count circular arrangements of this block plus the remaining persons. We also multiply by the number of internal permutations of the three persons within that block. Step-by-Step Solution: Step 1: Total number of circular arrangements of 14 distinct persons = (14 - 1)! = 13!.Step 2: For favourable arrangements, treat the three particular persons as one single block.Step 3: Then we effectively have 1 block + 11 other persons = 12 objects to arrange around the table.Step 4: The number of circular arrangements of these 12 objects = (12 - 1)! = 11!.Step 5: Inside the block, the three particular persons can be arranged among themselves in 3! = 6 ways.Step 6: Therefore, number of favourable arrangements = 11! * 3!.Step 7: Probability = favourable / total = (11! * 3!) / 13!.Step 8: Simplify 13! as 13 * 12 * 11!, so probability = 6 / (13 * 12) = 6 / 156 = 1 / 26. Verification / Alternative check: We could instead fix one of the three special persons as a reference and count ways to place the other two next to that person, but after taking into account the remaining persons and circular symmetry, we again obtain 1 / 26. Any correct method must yield the same simplified fraction, so the value 1 / 26 is consistent. Why Other Options Are Wrong: 3/26 and 5/26 correspond to larger favourable sets than are actually possible and would require more arrangements than exist. The fraction 1/13 doubles the correct probability, treating linear permutations instead of circular ones. The fraction 2/13 still overestimates the chance that three specific persons stay together and does not match the correct ratio of favourable to total circular arrangements. Common Pitfalls: A common mistake is to forget that the table is circular and to count all 14! linear arrangements, which overcounts each circular arrangement 14 times. Another error is to treat the three special persons as fixed in one spot rather than as a movable block, which distorts the count of favourable cases. Always remember to use (n - 1)! for circular arrangements and to multiply by internal permutations when treating several people as a single block. Final Answer: The required probability that the three particular persons always sit together is 1/26.

Question 6

In a leap year with 366 days, what is the probability that the calendar contains exactly 53 Saturdays and exactly 52 Sundays?

Accepted Answer

Introduction / Context: This question uses basic calendar facts combined with simple probability. A leap year has 366 days, which is 52 complete weeks plus 2 extra days. Every weekday must appear either 52 or 53 times. We are asked to find the probability that there are 53 Saturdays and only 52 Sundays in such a year. The problem reduces to understanding how the two extra days are distributed among the seven days of the week. Given Data / Assumptions: A leap year has 366 days. 366 days = 52 weeks + 2 days. Each day of the week occurs at least 52 times. Exactly two days of the week will occur 53 times. The first day of the year (1 January) is equally likely to be any of the 7 weekdays. Concept / Approach: If we write out the calendar, the 52 full weeks contribute 52 of each weekday. The remaining 2 days decide which weekdays get a 53rd occurrence. If 1 January is some weekday, the two extra days are that weekday and the next one in order. Therefore, the pair of extra days can be (Monday, Tuesday), (Tuesday, Wednesday), and so on, cycling through the week. We check which choices of starting day give exactly 53 Saturdays and only 52 Sundays. Step-by-Step Solution: Step 1: In a leap year, there are 52 Saturdays and 52 Sundays guaranteed from the 52 complete weeks.Step 2: There are 2 extra days, which are consecutive weekdays.Step 3: List the possible ordered pairs of these extra days depending on the starting day: (Mon, Tue), (Tue, Wed), (Wed, Thu), (Thu, Fri), (Fri, Sat), (Sat, Sun), (Sun, Mon).Step 4: To have 53 Saturdays, Saturday must appear among these extra days.Step 5: To have exactly 52 Sundays, Sunday must not appear among these extra days.Step 6: Among the listed pairs, the pair (Fri, Sat) includes Saturday but not Sunday. The pair (Sat, Sun) includes both Saturday and Sunday, which would give 53 Saturdays and 53 Sundays, not allowed.Step 7: Therefore, the only favourable case is when the extra days are (Fri, Sat), which occurs when 1 January is a Friday.Step 8: There are 7 equally likely choices for the day on which 1 January falls, and exactly 1 of these choices is favourable.Step 9: Hence, probability = 1 / 7. Verification / Alternative check: We can verify by reasoning directly on Saturdays. There are two ways for Saturday to be an extra day: the year starts on Friday or Saturday. However, when it starts on Saturday, the second extra day is Sunday, giving 53 Sundays as well. Since we require only 52 Sundays, only the Friday start works, reinforcing the answer 1 / 7. Why Other Options Are Wrong: 2/7 and 3/7 treat multiple starting days as favourable, including cases where Sunday also reaches 53 occurrences. The value 1/2 suggests that roughly half of all leap years satisfy the condition, which is far too large. The value 4/7 similarly overcounts many starts without checking the Sunday condition. Common Pitfalls: Learners sometimes stop as soon as they find all starts that give 53 Saturdays and forget to enforce the requirement that Sundays must stay at 52. Others misremember whether leap years add one day or two days beyond full weeks. Always remember that a leap year gives 52 full weeks and 2 extra days, and carefully check both conditions in probability questions involving days of the week. Final Answer: The probability that a leap year has exactly 53 Saturdays and 52 Sundays is 1/7.

Question 7

A person starts with 64 rupees and makes six bets in a game, where the probability of a win is equal to the probability of a loss. In each bet, the person stakes half of the money currently held. For each win, the stake is gained in addition to the …

Accepted Answer

Introduction / Context: This question tests understanding of percentage change and multiplicative growth rather than straightforward probability. The order of wins and losses might look important, but when each bet is for half the current capital, the final result depends only on how many wins and how many losses occur, not on their sequence. Recognising this simplifies the problem dramatically and avoids tedious case-by-case listing of different orders of wins and losses. Given Data / Assumptions: Initial capital is Rs. 64. The person makes 6 bets. Each bet is for half the money currently held. Three bets are wins and three bets are losses. Probability of a win equals probability of a loss, but we are already told the result pattern (3 wins, 3 losses), so we focus on the final amount. Concept / Approach: When the person bets half of the current capital, two different multiplicative factors apply. On a win, the person keeps the half that was not staked plus wins an equal amount to the stake. So the money is multiplied by 1.5. On a loss, the person loses the staked half and keeps only half of the capital, so the money is multiplied by 0.5. Since there are exactly three wins and three losses, the final capital is the initial capital multiplied by 1.5^3 and 0.5^3, in any order. Step-by-Step Solution: Step 1: After a win, factor = 1.5 (because capital becomes capital + half of capital = 1.5 times).Step 2: After a loss, factor = 0.5 (because capital is reduced to half).Step 3: The order of these factors does not matter, because multiplication is commutative.Step 4: Total multiplicative factor for three wins and three losses = 1.5^3 * 0.5^3.Step 5: Compute 1.5^3 = (3/2)^3 = 27/8.Step 6: Compute 0.5^3 = (1/2)^3 = 1/8.Step 7: Combined factor = (27/8) * (1/8) = 27/64.Step 8: Final capital = initial capital * combined factor = 64 * (27/64) = 27 rupees.Step 9: Net result compared to starting 64 rupees = final 27 rupees, so net loss = 64 - 27 = 37 rupees. Verification / Alternative check: To check, you can take a specific order such as win, win, win, loss, loss, loss and track the rupees step by step. After each win, multiply by 1.5, after each loss multiply by 0.5. Any order you choose will always give 27 rupees at the end, confirming that the order does not matter and that the multiplicative approach is correct. Why Other Options Are Wrong: A gain of Rs. 27 or Rs. 37 assumes the final amount is greater than the starting 64 rupees, which contradicts the fact that multiplying by 1.5 three times and by 0.5 three times yields an overall factor less than 1. A loss of Rs. 27 would correspond to a final amount of 37 rupees, which does not match the product 64 * 27 / 64. The option of no gain and no loss would mean the overall factor is 1, which is clearly not the case here. Common Pitfalls: Many learners mistakenly average the gains and losses and think that three wins and three losses cancel out. They forget that successive percentage changes do not simply cancel but combine multiplicatively. Another common error is to treat each win as plus a fixed amount, which is incorrect because each stake is half of the remaining money, not half of the original 64 rupees. Always convert percentage-style problems into multiplicative factors on the current amount and then multiply those factors in sequence. Final Answer: The person ends with Rs. 27, so the net outcome is a loss of Rs. 37 compared with the starting 64 rupees.

Question 8

Two letters are chosen at random without replacement from the word TIME. What is the probability that the selected letters are exactly T and M (in any order)?

Accepted Answer

Introduction / Context: This question introduces basic probability with simple combinations. We are given the four-letter word TIME and we randomly select two letters from it without replacement. The task is to find the probability that the chosen letters are exactly T and M, regardless of the order in which they are drawn. Because all letters in TIME are distinct, this turns into a straightforward counting problem using combinations. Given Data / Assumptions: The word TIME has 4 distinct letters: T, I, M and E. Two letters are selected at random without replacement. All 2-letter selections are equally likely. We want the selection that consists of the letters T and M, in any order. Concept / Approach: When order does not matter, we use combinations. The total number of ways to choose 2 letters from 4 distinct letters is given by 4C2. For the favourable outcomes, there is exactly one pair containing T and M together, because there is only one way to select these two specific letters when order is ignored. The probability is therefore the ratio of favourable combinations to total combinations. Step-by-Step Solution: Step 1: Count total ways to choose 2 letters from 4 distinct letters = 4C2.Step 2: Compute 4C2 = (4 * 3) / (2 * 1) = 6.Step 3: Count favourable selections. The pair we desire is {T, M}. There is exactly one combination containing T and M.Step 4: Number of favourable combinations = 1.Step 5: Probability = favourable / total = 1 / 6. Verification / Alternative check: We could list all 2-letter combinations: {T, I}, {T, M}, {T, E}, {I, M}, {I, E}, {M, E}. There are clearly 6 pairs and only one of them is {T, M}, which confirms that the probability is 1 out of 6. This listing approach is convenient when there are only a few possibilities and helps to cross-check the combination formula. Why Other Options Are Wrong: The fraction 1/4 might arise if someone incorrectly divides by the number of letters in the word instead of counting pairs. The value 1/8 and 1/12 are too small and come from confusing ordered pairs with unordered pairs or incorrectly squaring probabilities. The fraction 1/3 is too large and suggests that two different pairs are being treated as favourable by mistake. Only 1/6 correctly represents the ratio of the single favourable pair to the six possible pairs. Common Pitfalls: One common error is to treat the draws as ordered and to count TM and MT separately while also comparing to a total of 4 * 3 ordered pairs. Although this can be done correctly, many learners mix ordered and unordered thinking and arrive at inconsistent denominators. Another pitfall is to apply 2/4 * 1/3 reasoning, which does not directly represent the event that the unordered pair of letters is exactly T and M. The cleanest approach here is to use combinations and count unordered selections. Final Answer: The probability that the two chosen letters are T and M is 1/6.

Question 9

An urn contains 4 white balls, 6 black balls and 8 red balls. Three balls are drawn one after another without replacement. What is the probability that all three balls drawn are white?

Accepted Answer

Introduction / Context: This is a classic probability problem involving drawing balls from an urn without replacement. We are given the counts of white, black and red balls and asked to find the probability that all three balls drawn are white. Because the drawing is without replacement, the probabilities change after each draw, and we must either use combinations or sequential multiplication of conditional probabilities to find the answer. Given Data / Assumptions: White balls in the urn: 4. Black balls: 6. Red balls: 8. Total balls = 4 + 6 + 8 = 18. Three balls are drawn one after another without replacement. We want the event that all three drawn balls are white. Concept / Approach: There are two standard approaches. One uses combinations to count the number of ways to choose three balls from the urn and then restrict to white balls. The other uses the multiplication rule for successive probabilities, taking into account that after each white ball is drawn, both the number of white balls and the total number of balls decrease. Both approaches should lead to the same probability. Step-by-Step Solution (Combinational Method): Step 1: Total number of ways to choose any 3 balls from 18 = 18C3.Step 2: Compute 18C3 = (18 * 17 * 16) / (3 * 2 * 1) = 4896 / 6 = 816.Step 3: Number of favourable ways: choose 3 balls from the 4 white balls = 4C3.Step 4: Compute 4C3 = 4.Step 5: Probability = favourable / total = 4 / 816.Step 6: Simplify 4 / 816 = 1 / 204. Step-by-Step Solution (Sequential Probability): Step 1: Probability first ball is white = 4 / 18.Step 2: After drawing one white ball, white balls remaining = 3 and total balls = 17. Probability second ball is white = 3 / 17.Step 3: After drawing two white balls, white balls remaining = 2 and total balls = 16. Probability third ball is white = 2 / 16.Step 4: Multiply: overall probability = (4 / 18) * (3 / 17) * (2 / 16) = (4 * 3 * 2) / (18 * 17 * 16).Step 5: Simplify numerator and denominator to confirm that this also equals 1 / 204. Verification / Alternative check: Reducing (4 * 3 * 2) / (18 * 17 * 16) step by step gives 24 / (4896). Dividing numerator and denominator by 24, we get 1 / 204, matching the combinational method. This confirms that both approaches are consistent and that the probability is correctly simplified. Why Other Options Are Wrong: 5/204 and 13/204 are larger numerators but do not correspond to any meaningful count of favourable combinations. The fractions 3/68 and 1/68 have different denominators and would imply a completely different distribution of balls. Only 1/204 aligns with both the combination formula and the stepwise probability calculation. Common Pitfalls: Students sometimes forget that the draws are without replacement and treat each draw as 4/18, leading to (4/18)^3, which is incorrect. Others miscalculate 18C3 or 4C3, producing an incorrect denominator or numerator. Carefully distinguishing between with and without replacement and double-checking combination calculations are essential to avoid these mistakes. Final Answer: The probability that all three drawn balls are white is 1/204.

Question 10

Two brothers X and Y appear independently for a selection test. The probability that X is selected is 1/7 and the probability that Y is selected is 2/9. What is the probability that both brothers are selected?

Accepted Answer

Introduction / Context: This is a straightforward probability question involving independent events. Two brothers, X and Y, have given a test and each has a known probability of being selected. We are asked for the probability that both are selected. When two events are independent, the probability that both occur is the product of their individual probabilities. Given Data / Assumptions: Probability that X is selected = 1/7. Probability that Y is selected = 2/9. The selection outcomes for X and Y are assumed to be independent. We want the probability that X is selected and Y is selected. Concept / Approach: For independent events A and B, the probability that both occur is P(A and B) = P(A) * P(B). Here, event A is "X is selected" and event B is "Y is selected". Because the outcome for one brother does not affect the outcome for the other, we can multiply the two probabilities directly to obtain the required result. Step-by-Step Solution: Step 1: Let A be the event that X is selected, so P(A) = 1/7.Step 2: Let B be the event that Y is selected, so P(B) = 2/9.Step 3: Since the events are independent, P(A and B) = P(A) * P(B).Step 4: Compute P(A and B) = (1/7) * (2/9) = 2 / 63.Step 5: Therefore, the probability that both brothers are selected is 2/63. Verification / Alternative check: We can also think in terms of a simple grid of outcomes: X may be selected or not selected, and Y may be selected or not selected. The probability for any particular pair of outcomes is the product of the corresponding individual probabilities. The desired outcome is (selected, selected), which has probability 1/7 * 2/9 = 2/63, matching our earlier calculation. Why Other Options Are Wrong: 1/63 would correspond to incorrectly multiplying 1/7 and 1/9 or misreading the selection probability of Y. The fraction 1/14 is larger than 2/63 and might come from adding probabilities instead of multiplying. The value 1/9 ignores the selection probability of X, and 4/63 would require either both probabilities to be larger or incorrect multiplication. Only 2/63 correctly reflects the product of 1/7 and 2/9. Common Pitfalls: A typical mistake is to add probabilities (1/7 + 2/9) instead of multiplying them, which would give the probability that at least one is selected only if we also subtracted the overlap. Another mistake is to treat the events as mutually exclusive, which they are not, because both brothers can be selected simultaneously. Always check whether events are independent or mutually exclusive before choosing the appropriate rule. Final Answer: The probability that both X and Y are selected is 2/63.

Question 11

Two cards are drawn at random without replacement from a well shuffled standard deck of 52 cards. What is the probability that one of the cards is a club (any club) and the other card is a king of a non club suit?

Accepted Answer

Introduction / Context: This question involves card probability with two different types of cards: any club and a king from a non club suit. The deck is standard with 52 cards, 13 in each suit and 4 kings in total. Because the cards are drawn without replacement, outcomes must be counted using combinations of two card hands rather than independent single card probabilities alone. We want exactly one club and one king of a non club suit, so the king of clubs is not allowed to count as both items at once. Given Data / Assumptions: Standard deck = 52 cards. Clubs in the deck = 13 cards. Kings in the deck = 4 cards. King of clubs is excluded from the king choice, because we require a club card and a separate non club king. Two cards are drawn at random without replacement. The event of interest is: one card is a club (any rank) and the other is a king of hearts, diamonds or spades. Concept / Approach: We work with unordered pairs of cards because in a random two card hand, the order of drawing does not matter for the final combination. The total number of possible two card hands is 52C2. Favourable hands consist of one club and one non club king. Since the sets of clubs and non club kings are disjoint, we can count the favourable pairs simply by multiplying the number of possibilities for each card type. Step-by-Step Solution: Step 1: Total number of two card hands = 52C2.Step 2: Compute 52C2 = (52 * 51) / (2 * 1) = 1326.Step 3: Count the number of club cards: there are 13 clubs in the deck.Step 4: Count the number of kings that are not clubs. There are 4 kings in total and exactly 1 king of clubs, so non club kings = 4 - 1 = 3.Step 5: To form a favourable hand, choose 1 club and 1 non club king. Because the two groups are disjoint, number of favourable hands = 13 * 3 = 39.Step 6: Probability = favourable / total = 39 / 1326.Step 7: Simplify 39 / 1326 by dividing numerator and denominator by 3 to obtain 13 / 442. Verification / Alternative check: We could also compute the probability using an ordered approach and then adjust. For example, choose the club first and then the non club king, or vice versa. The probability for club then non club king is (13/52) * (3/51). The probability for non club king then club is (3/52) * (13/51). Adding these gives (13 * 3 / (52 * 51)) * 2 = 78 / 2652 = 39 / 1326 = 13 / 442, confirming the combination method. Why Other Options Are Wrong: The value 29/442 arises from counting an incorrect number of favourable pairs. The fractions 1/26 and 1/13 are much larger probabilities and correspond to more frequent events, such as drawing any king or drawing a club without restricting the second card. The value 3/52 is closer to the probability of drawing a king in one draw, but it does not incorporate the specific requirement of having both card types simultaneously. Only 13/442 is consistent with correct counting of two card hands. Common Pitfalls: One pitfall is to allow the king of clubs as part of the event, which would double count cases where a single card fulfils both conditions. Another is to mix up ordered and unordered reasoning, for example counting 39 favourable ordered outcomes but dividing by 52C2. Both ordered and unordered methods work, but the counting and denominator must match. Always carefully interpret the wording: here, the club and the king must be distinct cards, and the king must come from a non club suit. Final Answer: The probability that one card is a club and the other is a king of a non club suit is 13/442.

Question 12

A bag contains 3 red balls, 5 yellow balls and 7 pink balls. If one ball is drawn at random from the bag, what is the probability that the ball drawn is either pink or red?

Accepted Answer

Introduction / Context: This is a simple one step probability problem involving a single draw from a bag containing coloured balls. The event of interest is that the ball drawn is either pink or red. We are given the exact counts of each colour, and the drawing is at random, so each ball has the same chance of being chosen. Such questions test basic understanding of favourable outcomes divided by total outcomes. Given Data / Assumptions: Red balls: 3. Yellow balls: 5. Pink balls: 7. Total balls = 3 + 5 + 7 = 15. One ball is drawn at random. We need the probability that the ball is pink or red. Concept / Approach: The probability of an event is defined as the number of favourable outcomes divided by the total number of equally likely outcomes. Here, the outcomes are the specific balls in the bag. A favourable outcome occurs when the chosen ball is either red or pink. Because these colour groups do not overlap and there is no restriction about yellow, we simply add the counts of red and pink balls to find the number of favourable outcomes. Step-by-Step Solution: Step 1: Compute total number of balls = 3 red + 5 yellow + 7 pink = 15.Step 2: Count favourable balls: these are balls that are either red or pink.Step 3: Number of red balls = 3.Step 4: Number of pink balls = 7.Step 5: Total favourable balls = 3 + 7 = 10.Step 6: Probability = favourable / total = 10 / 15.Step 7: Simplify 10 / 15 by dividing numerator and denominator by 5 to get 2 / 3. Verification / Alternative check: We can verify by computing complementary probabilities. The only colour that does not count as favourable is yellow. There are 5 yellow balls, so the probability of drawing yellow is 5 / 15 = 1 / 3. The probability of drawing red or pink is therefore 1 - 1 / 3 = 2 / 3, which matches the direct calculation and confirms the correctness of the answer. Why Other Options Are Wrong: The fraction 3/5 would correspond to 9 favourable balls out of 15, which is not correct because there are 10. The value 3/8 does not relate to the actual counts and would imply a total of 8 outcomes. The probability 1/8 is far too small and ignores the fact that most balls are either pink or red. The fraction 7/15 would be correct if only pink balls were favourable, but here both pink and red count, so we must include both colours. Common Pitfalls: A common mistake is to miscount total balls or to forget one of the colours when computing favourable outcomes. Some learners overcomplicate the problem by trying to use combinations, even though this is a single simple selection. Always remember that when each individual object is equally likely, you can often work directly with counts rather than more complex formulas. Final Answer: The probability that the ball drawn is either pink or red is 2/3.

Question 13

When two fair dice are thrown simultaneously, what is the probability that the sum of the two numbers obtained is less than 12?

Accepted Answer

Introduction / Context: This question deals with the sample space of two fair dice. Each die has 6 faces and all ordered pairs of outcomes are equally likely. We are asked for the probability that the sum of the two numbers is less than 12. The maximum possible sum when two dice are thrown is 12, which occurs only when both dice show 6. Recognising this allows us to solve the problem very quickly using a complement argument. Given Data / Assumptions: Two fair six faced dice are thrown at the same time. Each die can show an integer from 1 to 6. All 36 ordered pairs (1,1) to (6,6) are equally likely. We are interested in the event that the sum of the two numbers is less than 12. Concept / Approach: Instead of counting all the sums less than 12 directly, it is easier to compute the probability of the complement event and subtract from 1. The only way the sum can be 12 is if both dice show 6. Thus, P(sum = 12) is straightforward to compute. Then P(sum less than 12) = 1 - P(sum = 12). This is a standard example of using the complement rule in probability to simplify counting. Step-by-Step Solution: Step 1: Total number of possible ordered outcomes when two dice are thrown = 6 * 6 = 36.Step 2: The maximum sum is 12, which occurs only when the first die is 6 and the second die is 6.Step 3: There is exactly 1 ordered outcome that gives a sum of 12, namely (6, 6).Step 4: Probability that the sum is 12 = 1 / 36.Step 5: The event that the sum is less than 12 is the complement of the event that the sum is 12, because every other outcome has sum from 2 to 11.Step 6: Therefore, P(sum less than 12) = 1 - P(sum = 12) = 1 - 1 / 36.Step 7: Compute 1 - 1 / 36 = 35 / 36. Verification / Alternative check: If desired, we could check explicitly that all sums from 2 to 11 are achievable and that at least one outcome leads to each of these sums. For any given sum from 2 to 11, you can find at least one ordered pair (such as (1,1) for 2, (1,2) for 3, and so on). Since 36 total outcomes exist and exactly one outcome gives sum 12, the other 35 outcomes must correspond to sums less than 12, validating the complement method. Why Other Options Are Wrong: The fractions 17/36 and 15/36 correspond to much smaller subsets of outcomes and would imply that many sums greater than or equal to some value have been excluded incorrectly. The value 1/36 is actually the probability of getting a sum of exactly 12, not less than 12. The fraction 5/36 is also much too small and would mean that only a few outcomes satisfy the condition, which is not true here. Common Pitfalls: Some learners mistakenly count only a subset of sums or miscount the total sample space, forgetting that the dice outcomes are ordered pairs. Others confuse the event "sum less than 12" with "sum less than or equal to some smaller number" and undercount favourable outcomes. Using the complement rule is a powerful and safe approach whenever the complement event has a very simple structure, as in this case. Final Answer: The probability that the sum of the two dice is less than 12 is 35/36.

Question 14

A fair coin is tossed 5 times in succession. What is the probability that the number of heads obtained is an odd number (that is, 1, 3 or 5)?

Accepted Answer

Introduction / Context: This question examines the binomial distribution for repeated coin tosses. We toss a fair coin five times and want the probability that the number of heads is odd. The possible odd counts in 5 tosses are 1, 3 or 5. There is a useful symmetry result for fair coins: the probability of an odd number of heads equals the probability of an even number of heads, so each is 1/2. We can use symmetry or compute directly using binomial coefficients. Given Data / Assumptions: A fair coin is tossed 5 times. Each toss has two possible outcomes: head or tail. The probability of head in each toss is 1/2, and tail is 1/2. We seek the probability of getting an odd number of heads, namely 1, 3 or 5 heads. Concept / Approach: The total number of possible outcomes for 5 tosses is 2^5. The number of ways to obtain exactly k heads is given by 5Ck. Using the binomial distribution, P(k heads) = 5Ck * (1/2)^5. Since the coin is fair, we can either calculate P(1 head) + P(3 heads) + P(5 heads) directly or use the symmetry that the probability of an odd number of heads equals the probability of an even number of heads, which must be 1/2 each. Step-by-Step Solution (Direct Method): Step 1: Total number of outcomes = 2^5 = 32.Step 2: Probability of exactly 1 head = 5C1 * (1/2)^5 = 5 * (1/32) = 5/32.Step 3: Probability of exactly 3 heads = 5C3 * (1/2)^5 = 10 * (1/32) = 10/32.Step 4: Probability of exactly 5 heads = 5C5 * (1/2)^5 = 1 * (1/32) = 1/32.Step 5: Probability of an odd number of heads = 5/32 + 10/32 + 1/32 = 16/32 = 1/2. Verification / Symmetry Argument: For a fair coin, the distribution of heads and tails is symmetric. If we flip the coin 5 times, every outcome with k heads has a corresponding outcome with 5 - k heads obtained by swapping heads and tails in each toss. Thus, the probability of even heads equals the probability of odd heads. Since these two cases cover all possibilities, each must have probability 1/2. This symmetry matches our detailed calculation of 16 favourable outcomes out of 32. Why Other Options Are Wrong: The value 1/3 would correspond to about 10.67 favourable outcomes out of 32, which is not an integer count. The value 2/3 is larger than 1/2 and would require more odd head combinations than are actually possible. The value 1 would mean that an odd number of heads occurs every time, which is clearly not true. The fraction 3/8 is 12/32, less than the correct 16 favourable outcomes and misses part of the distribution. Common Pitfalls: Some learners forget to include all odd cases (for example, they might account for 1 and 3 heads but forget 5 heads). Others miscalculate the binomial coefficients 5C3 or 5C1. A third common issue is treating the events of 1 head, 3 heads and 5 heads as overlapping, which they are not. Using the binomial formula carefully and checking the symmetry between heads and tails helps avoid these mistakes. Final Answer: The probability that heads appear an odd number of times in five tosses is 1/2.

Question 15

A five digit number is formed using the digits 0, 1, 2, 3 and 4 without repetition. The first digit cannot be 0. What is the probability that the number formed is divisible by 5?

Accepted Answer

Introduction / Context: This question is a blend of basic number theory and counting methods. We form five digit numbers using the digits 0, 1, 2, 3 and 4 without repetition, under the restriction that the first digit of the number cannot be 0. We are asked to find the probability that such a number is divisible by 5. A number is divisible by 5 if and only if its last digit is 0 or 5; in this problem, the digit 5 is not available, so the last digit must be 0. Given Data / Assumptions: Available digits: 0, 1, 2, 3 and 4. Each digit can be used at most once (no repetition). The first digit of the number cannot be 0. The number formed must be five digits long. We want the probability that the number is divisible by 5. Concept / Approach: We first count the total number of valid five digit numbers we can form under the no repetition and non zero leading digit constraint. Then we count the number of such numbers that are divisible by 5. Since 5 is not among the allowed digits, divisibility by 5 is possible only when the last digit is 0. The required probability is the ratio of the favourable count to the total count. Step-by-Step Solution (Total Numbers): Step 1: Total permutations of the 5 distinct digits without any restriction = 5! = 120.Step 2: Count invalid permutations where the first digit is 0. If 0 is fixed at the first position, the remaining 4 digits can be arranged in 4! = 24 ways.Step 3: Therefore, valid five digit numbers (with first digit not 0) = 120 - 24 = 96. Step-by-Step Solution (Favourable Numbers): Step 1: For divisibility by 5, last digit must be 0, because 5 is not one of the digits.Step 2: Fix 0 in the last position. The remaining four positions (from left to right) must be filled by the digits 1, 2, 3 and 4.Step 3: Since these four digits are all non zero, any arrangement of them in the first four positions yields a valid five digit number.Step 4: Number of ways to arrange 1, 2, 3 and 4 in the first four positions = 4! = 24.Step 5: Therefore, the number of favourable five digit numbers divisible by 5 = 24.Step 6: Probability = favourable / total = 24 / 96 = 1 / 4. Verification / Alternative check: We can double check by noticing that all valid five digit numbers (with these digits and a non zero first digit) are equally likely if we choose a random arrangement. Since the last digit can equally likely be any of the five digits subject to the non zero first digit constraint, and only one choice for the last digit (0) gives divisibility by 5, the fraction of such numbers is effectively 1 out of 4 possible non leading positions, aligning with the more explicit counting result of 1 / 4. Why Other Options Are Wrong: The probability 1/5 would be correct only if any digit could appear in the last place with equal likelihood and digits 0 and 5 were available. However, 5 is missing here. The values 2/5, 3/5 and 1/2 are too large and imply that a much higher fraction of numbers are divisible by 5 than is actually the case. Only 1/4 matches the exact counting of valid and favourable permutations. Common Pitfalls: Students sometimes forget to remove the invalid numbers with leading 0 when counting the total. Others mistakenly assume that both 0 and 5 are available as last digits or that the five digit number can start with 0. Carefully handling the first digit restriction and the divisibility rule ensures a correct solution. Final Answer: The probability that the five digit number formed is divisible by 5 is 1/4.

Question 16

In probability theory, what is the numerical value of the probability of an impossible event?

Accepted Answer

Introduction / Context: This is a conceptual question about the basic axioms of probability. An impossible event is one that can never occur under the given experimental setup, no matter how many times the experiment is repeated. Understanding the correct probability value for such an event is fundamental to mastering probability theory and interpreting probability values correctly. Given Data / Assumptions: We are working within standard probability theory. An event is called impossible if it has no possible outcomes in the sample space. We must choose the correct numerical value for the probability of an impossible event from the given options. Concept / Approach: By the axioms of probability, the probability of any event E satisfies 0 ≤ P(E) ≤ 1. The probability of the sure (certain) event is 1, and the probability of the empty event, or impossible event, must be 0. The empty event has no favourable outcomes, so the ratio of favourable outcomes to total outcomes is 0 divided by the total, which is 0. This is built into every standard definition of a probability measure. Step-by-Step Reasoning: Step 1: Consider an experiment with a finite sample space S and let n(S) represent the number of all possible outcomes.Step 2: For a given event E, let n(E) be the number of favourable outcomes for that event.Step 3: The basic definition of probability in this finite, equally likely case is P(E) = n(E) / n(S).Step 4: For an impossible event, E is the empty set, and hence n(E) = 0.Step 5: Therefore, P(E) = 0 / n(S) = 0.Step 6: This aligns with the general axiom that the probability of the empty set is always 0. Verification / Alternative check: In measure theoretic terms, probability is a function P defined on events such that P(S) = 1 and P is countably additive. The empty set has measure zero in all standard probability spaces because it cannot be decomposed into any non empty events and carries no weight in sums. Any other value, such as a negative number or a positive number, would contradict the property that probabilities must lie between 0 and 1 and that P(S) = 1 while P of disjoint events adds to the total. Why Other Options Are Wrong: The value 1 is the probability of a certain event, not an impossible event. A negative value such as -1 is not allowed because probabilities cannot be less than 0. The value 0.1 or 0.5 would imply that the supposedly impossible event actually has some chance of occurring, which contradicts the definition. Only 0 is consistent with both the intuitive and formal definitions of an impossible event. Common Pitfalls: Some learners confuse an impossible event with a very unlikely event. An event with very small probability, such as 0.001, is unlikely but still possible; an impossible event truly cannot happen and therefore must have probability 0. Always distinguish between improbable events (small but positive probabilities) and impossible events (exactly 0 probability). Final Answer: The probability of an impossible event is 0.

Question 17

Three fair dice are thrown simultaneously. What is the probability that at least one of the dice shows a six?

Accepted Answer

Introduction / Context: This question uses the complement rule in probability for repeated independent trials. We throw three fair dice at the same time and want the probability that at least one die shows the number six. Directly counting arrangements with one or more sixes can be messy. Instead, it is much easier to calculate the probability that no die shows a six and subtract this from 1 to obtain the probability of at least one six. Given Data / Assumptions: Three fair six faced dice are thrown simultaneously. Each die has faces numbered 1 to 6. The dice are independent, and each face on each die has probability 1/6. We are interested in the event that at least one die shows the face 6. Concept / Approach: The complement of the event "at least one six" is the event "no six appears". If we can find the probability that none of the three dice shows a six, we can subtract this probability from 1. This method significantly reduces counting errors and is standard for "at least one" type problems in probability. Step-by-Step Solution: Step 1: For a single fair die, probability that it does not show a six is 5/6.Step 2: Because the dice are independent, the probability that all three dice show numbers that are not six is (5/6)^3.Step 3: Compute (5/6)^3 = (5 * 5 * 5) / (6 * 6 * 6) = 125 / 216.Step 4: This is the probability of the complement event (no six at all).Step 5: Probability of at least one six = 1 - probability of no six.Step 6: Therefore, P(at least one six) = 1 - 125 / 216 = (216 - 125) / 216 = 91 / 216. Verification / Alternative check: We can also check by computing probability of exactly one six, exactly two sixes and exactly three sixes and then summing them. For example, P(exactly one six) = 3 * (1/6) * (5/6)^2, P(exactly two sixes) = 3 * (1/6)^2 * (5/6), and P(exactly three sixes) = (1/6)^3. Summing these three values also gives 91 / 216 after simplification. This verifies that the complement based calculation is correct. Why Other Options Are Wrong: The value 1/36 is much too small and does not reflect the many ways at least one six can occur on three dice. The fraction 13/216 is too small and could correspond to a poorly calculated partial event. The value 125/216 is actually the probability of the complement event (no six on any die), not the desired event. The fraction 5/36 is also incorrect and does not correspond to any natural counting pattern for this experiment. Common Pitfalls: A common mistake is to treat "at least one six" as equivalent to "exactly one six" and only count those cases. Another frequent error is misusing addition or not recognising independence, leading to wrong exponents in the calculation. Using the complement rule and clearly identifying the no six event as (5/6)^3 is the safest route. Final Answer: The probability that at least one of the three dice shows a six is 91/216.

Question 18

Ajay and his wife Reshmi appear independently for two vacancies in the same post. The probability that Ajay is selected is 1/7 and the probability that Reshmi is selected is 1/5. What is the probability that exactly one of them is selected?

Accepted Answer

Introduction / Context: This probability question focuses on the idea of exactly one of two independent events occurring. Ajay and Reshmi both apply for two vacancies, and we know their individual selection probabilities. We must find the probability that exactly one of them is selected, meaning either Ajay alone or Reshmi alone, but not both together. This is an example of the event known as the symmetric difference of two independent events. Given Data / Assumptions: Probability that Ajay is selected = 1/7. Probability that Reshmi is selected = 1/5. Their selection events are independent. We need the probability that exactly one is selected. Concept / Approach: Let A be the event that Ajay is selected and R be the event that Reshmi is selected. The event "exactly one is selected" can be written as (A and not R) or (R and not A). Because A and R are independent, the probabilities of combinations such as A and not R can be found by multiplying the probabilities of the constituent events. We then add the mutually exclusive events A and not R, and R and not A. Step-by-Step Solution: Step 1: Let P(A) = 1/7 and P(R) = 1/5.Step 2: Probability that Ajay is selected and Reshmi is not selected = P(A) * P(not R) = (1/7) * (1 - 1/5) = (1/7) * (4/5) = 4/35.Step 3: Probability that Reshmi is selected and Ajay is not selected = P(R) * P(not A) = (1/5) * (1 - 1/7) = (1/5) * (6/7) = 6/35.Step 4: These two events are mutually exclusive because they cannot happen at the same time.Step 5: Probability that exactly one is selected = 4/35 + 6/35 = 10/35.Step 6: Simplify 10/35 by dividing numerator and denominator by 5 to get 2/7. Verification / Alternative check: An alternative is to use the formula P(exactly one) = P(A) + P(R) - 2 * P(A and R). Since P(A and R) = (1/7) * (1/5) = 1/35, we obtain P(exactly one) = 1/7 + 1/5 - 2 * (1/35) = 5/35 + 7/35 - 2/35 = 10/35 = 2/7. This confirms the result obtained from direct case analysis. Why Other Options Are Wrong: The fraction 2/35 is actually P(A and R), the probability that both are selected, not that exactly one is selected. The value 5/7 is much too large and would imply that in most scenarios exactly one is chosen, which is not supported by the given probabilities. The values 1/5 and 1/7 correspond to the individual selection probabilities and do not capture the combined event correctly. Common Pitfalls: A common mistake is to compute P(A) + P(R) and stop there, forgetting to subtract the probability that both are selected, which is counted twice in that sum. Another error is to treat the events as mutually exclusive, which they are not. Always remember that "exactly one" requires including two separate cases and excluding the overlap where both are selected. Final Answer: The probability that exactly one of Ajay or Reshmi is selected is 2/7.

Question 19

A box contains 10 bulbs, of which exactly 3 are defective and the remaining 7 are good. If a random sample of 5 bulbs is drawn without replacement, what is the probability that the sample contains no defective bulbs?

Accepted Answer

Introduction / Context: This question is a typical application of combinations in probability, specifically the hypergeometric distribution. We have a finite population of bulbs with some defective and some good bulbs. A sample of fixed size is drawn without replacement, and we are asked for the probability that no defective bulbs appear in the sample. This is a common model for quality control and acceptance sampling problems. Given Data / Assumptions: Total bulbs in the box = 10. Defective bulbs = 3. Good bulbs = 10 - 3 = 7. A random sample of 5 bulbs is drawn without replacement. We want the probability that all 5 bulbs in the sample are good (no defective bulbs). Concept / Approach: We treat every possible 5 bulb sample as equally likely. The total number of possible samples is the number of combinations of 10 bulbs taken 5 at a time. The favourable samples are those that contain only good bulbs, which means all 5 are chosen from the 7 good bulbs. The required probability is favourable combinations divided by total combinations. Step-by-Step Solution: Step 1: Total number of ways to choose any 5 bulbs from 10 = 10C5.Step 2: Compute 10C5 = (10 * 9 * 8 * 7 * 6) / (5 * 4 * 3 * 2 * 1) = 252.Step 3: Favourable samples must contain only good bulbs. There are 7 good bulbs.Step 4: Number of ways to choose 5 good bulbs from 7 = 7C5.Step 5: Compute 7C5 = 7C2 = (7 * 6) / (2 * 1) = 21.Step 6: Probability that a random sample of 5 has no defective bulbs = favourable / total = 21 / 252.Step 7: Simplify 21 / 252 by dividing numerator and denominator by 21 to get 1 / 12. Verification / Alternative check: We can confirm by using the sequential probability approach. The probability that the first bulb is good is 7/10, the second is good given the first was good is 6/9, then 5/8, then 4/7 and finally 3/6. The product (7/10) * (6/9) * (5/8) * (4/7) * (3/6) simplifies to 21/252, which again equals 1/12. This matches the combination method and verifies the result. Why Other Options Are Wrong: The fractions 5/12 and 7/12 are much larger and would suggest that most samples are completely free of defectives, which is not consistent with having 3 defective bulbs among 10. The value 3/14 is also incorrect and arises from incomplete simplification or wrong counting. The value 1/4 is closer but still not equal to the exact fraction 1/12 that follows from both combinational and sequential reasoning. Common Pitfalls: One common error is to treat the draws as with replacement and calculate (7/10)^5, which ignores the decreasing pool of good bulbs. Another mistake is to accidentally choose some defective bulbs when counting favourable outcomes, or to miscompute 10C5 or 7C5. Always verify combination calculations and keep track of whether the sampling is with or without replacement. Final Answer: The probability that the five bulb sample contains no defective bulbs is 1/12.

Question 20

Out of 17 applicants for a job (8 boys and 9 girls), two persons are to be selected at random. What is the probability that at least one of the selected persons will be a girl?

Accepted Answer

Introduction / Context: This question illustrates the use of combinations and the complement rule in probability. We are selecting two persons at random from a group of applicants consisting of boys and girls, and we want the probability that there is at least one girl among those selected. Often, it is easier to find the probability of the complementary event (no girl selected) and subtract this from 1 to get the final answer. Given Data / Assumptions: Total applicants = 17. Boys = 8. Girls = 9. Two persons are chosen at random. We are interested in the event that at least one of the selected persons is a girl. Concept / Approach: Instead of counting combinations that contain at least one girl directly, we consider the complement event: both selected persons are boys. The probability of at least one girl is then 1 minus the probability that both selected are boys. This method keeps the counting simple and reliable, especially when there are many ways to include at least one girl but only a few ways to include no girls. Step-by-Step Solution: Step 1: Total number of ways to choose 2 persons out of 17 = 17C2.Step 2: Compute 17C2 = (17 * 16) / (2 * 1) = 136.Step 3: For the complement event (no girl selected), both selected persons must be boys.Step 4: Number of ways to choose 2 boys out of 8 = 8C2.Step 5: Compute 8C2 = (8 * 7) / (2 * 1) = 28.Step 6: Probability that both selected are boys = 28 / 136.Step 7: Simplify 28 / 136 by dividing numerator and denominator by 4 to get 7 / 34.Step 8: Probability that at least one of the selected persons is a girl = 1 - 7 / 34 = (34 - 7) / 34 = 27 / 34. Verification / Alternative check: We can double check by computing probabilities directly. The probability that the first selected person is a boy is 8 / 17, and if this happens, the probability that the second is also a boy is 7 / 16. Thus, P(both boys) = (8 / 17) * (7 / 16) = 56 / 272 = 7 / 34. Taking 1 minus this value again gives 27 / 34 for at least one girl, confirming the combination based method. Why Other Options Are Wrong: The value 19/34 or 25/34 would arise from incorrect subtraction or miscounting the all boy combinations. The fraction 20/34 simplifies to 10/17 and does not correspond to any correct counting argument. The value 7/34 is the probability of selecting two boys and hence represents the complement event, not the event with at least one girl. Common Pitfalls: Some learners accidentally count the cases with exactly one girl and exactly two girls separately and make arithmetic errors when adding. Others forget that the group contains more girls than boys and guess that the probability is close to 1/2. The complement approach, focusing on the simpler event of choosing two boys, is usually the safest and fastest route. Final Answer: The probability that at least one of the two selected persons is a girl is 27/34.

Probability Questions