Question 1

A bag contains 5 red smileys, 6 yellow smileys and 3 green smileys. If two smileys are picked at random without replacement, what is the probability that both smileys picked are red or both are green?

Accepted Answer

Introduction / Context: This question requires combining probabilities of two mutually exclusive events in a single draw without replacement. We have smileys of three different colours in a bag, and we pick two smileys. The event of interest is that either both picked smileys are red or both are green. The two sub events cannot happen at the same time in a two smiley draw, so their probabilities can be added directly. Given Data / Assumptions: Red smileys = 5. Yellow smileys = 6. Green smileys = 3. Total smileys = 5 + 6 + 3 = 14. Two smileys are picked at random without replacement. We want the probability that both are red or both are green. Concept / Approach: We treat this as drawing an unordered pair of smileys from the bag. The total number of possible pairs is 14C2. Favourable outcomes are of two types: both red or both green. We count each type separately (5C2 for red pairs and 3C2 for green pairs), then sum and divide by the total number of pairs. Because a pair cannot be simultaneously all red and all green, the two events are mutually exclusive and we can add their probabilities. Step-by-Step Solution: Step 1: Total number of ways to choose 2 smileys from 14 = 14C2.Step 2: Compute 14C2 = (14 * 13) / (2 * 1) = 91.Step 3: Number of ways to choose 2 red smileys from 5 = 5C2 = (5 * 4) / 2 = 10.Step 4: Number of ways to choose 2 green smileys from 3 = 3C2 = (3 * 2) / 2 = 3.Step 5: Total favourable ways (both red or both green) = 10 + 3 = 13.Step 6: Probability = favourable / total = 13 / 91.Step 7: Simplify 13 / 91 by dividing numerator and denominator by 13 to get 1 / 7. Verification / Alternative check: We can verify using sequential probabilities. For both red: probability first is red = 5/14, then second is red given first was red = 4/13, so P(both red) = (5/14) * (4/13) = 20 / 182 = 10 / 91. For both green: probability first is green = 3/14, then second green = 2/13, so P(both green) = (3/14) * (2/13) = 6 / 182 = 3 / 91. Summing gives (10 / 91) + (3 / 91) = 13 / 91 = 1 / 7, confirming our combination method. Why Other Options Are Wrong: The fraction 3/7 is three times the correct probability and would overstate how often we get two reds or two greens. The fraction 7/91 is half the correct probability and could result from counting only red pairs or only green pairs. The value 0 would be correct only if it were impossible to get two smileys of the same colour, which is not the case. The value 13/91 is equivalent to the unsimplified fraction but is less reduced than 1/7; the simplest correct form is 1/7. Common Pitfalls: Students sometimes forget to simplify fractions and may stop at 13/91, not realising that it reduces to 1/7. Others accidentally include red and green mixed pairs in the favourable set, which does not satisfy the condition of both being of the same specified colour. Carefully distinguishing between the two favourable categories and using combinations correctly avoids these errors. Final Answer: The probability that both smileys picked are red or both are green is 1/7.

Question 2

If the difference between the mode and median is 2, then the difference between the median and mean is:

Accepted Answer

1

Question 3

A fair coin is tossed once. If the coin shows a tail, then a fair six faced die is tossed once. Given that the first toss of the coin results in a tail (so the die is definitely tossed), what is the probability that the die shows a number greater th…

Accepted Answer

Introduction / Context: This question links a coin toss to a subsequent die roll. The die is tossed only if the first coin toss shows a tail. We are told that the first toss results in a tail, so the die is definitely rolled. The problem then reduces to a simple probability question about a fair die: what is the chance that the number obtained is greater than 4? This tests understanding of conditional probability in a context where the condition fixes which part of the experiment actually happens. Given Data / Assumptions: A fair coin is tossed once. If the coin shows a tail, a fair six faced die is tossed once. We are given that the first coin toss resulted in a tail, so the die is certainly rolled. The die is fair, with faces numbered 1 to 6. We want the probability that the die shows a number greater than 4. Concept / Approach: Once we know that the first toss was a tail, the event concerning the coin is already fixed and has probability 1 in the conditional world we are considering. The remaining random part of the experiment is the die roll. For a fair six faced die, the outcomes greater than 4 are 5 and 6. Each face has equal probability 1/6, so we can directly count favourable outcomes and divide by the total number of outcomes for the die. Step-by-Step Solution: Step 1: Possible outcomes of a fair die are the numbers 1, 2, 3, 4, 5 and 6.Step 2: The event of interest is that the die shows a number greater than 4.Step 3: Numbers greater than 4 on the die are 5 and 6.Step 4: Number of favourable outcomes for the die = 2 (namely 5 and 6).Step 5: Total possible outcomes when the die is rolled = 6.Step 6: Probability that the die shows a number greater than 4 = favourable / total = 2 / 6.Step 7: Simplify 2 / 6 by dividing numerator and denominator by 2 to get 1 / 3. Verification / Alternative check: If we were to consider the entire experiment including the coin, we would first note that the die is rolled only in half of all experimental runs (when the coin shows tail). However, the question conditions on the event that the first toss is tail, effectively restricting attention to exactly those runs. Within that restricted set, the die behaviour is unaffected and remains uniformly random over its six faces. Therefore, the unconditional structure does not change the conditional probability 1/3 for outcomes 5 or 6. Why Other Options Are Wrong: The value 2/3 would be correct if four outcomes of the die were considered favourable, which is not the case here. The probability 2/5 is associated with a five outcome sample space, which does not fit a standard die. The fraction 4/15 and 1/6 both arise from mixing coin and die stages incorrectly without recognising the conditioning; they no longer apply once we know that the die has indeed been rolled. Common Pitfalls: Some learners mistakenly include the coin toss probability after conditioning, for example, multiplying 1/2 for tail with 1/3 for the die and obtaining 1/6, which would be the joint probability of tail and a number greater than 4 without conditioning. However, the question already tells us that a tail occurred, so we must work within that conditional universe. Always read the phrase "given that" as a clear signal to condition on an event, which may simplify what remains. Final Answer: Given that the first coin toss is a tail, the probability that the die shows a number greater than 4 is 1/3.

Question 4

A room contains 3 brown chairs, 5 black chairs and 4 white chairs. Two chairs are picked at random without replacement and placed in the lawn. What is the probability that none of the chairs picked is white?

Accepted Answer

Introduction / Context: This question tests probability with selections from a finite set of objects of different types. We have chairs of three different colours and we pick two chairs at random without replacement. The requirement is that none of the chairs picked is white, so both selected chairs must be non white. This can be solved conveniently using combinations. Given Data / Assumptions: Brown chairs = 3. Black chairs = 5. White chairs = 4. Total chairs in the room = 3 + 5 + 4 = 12. Two chairs are chosen without replacement. We want the probability that neither selected chair is white (both chairs are from brown or black). Concept / Approach: We treat the selection as a two chair combination from the total chairs. The total number of ways to choose any two chairs from 12 is 12C2. The favourable selections are those in which both chairs come from the non white group. The non white chairs are the brown and black ones. We count the ways of choosing 2 chairs from this non white group and divide by the total number of combinations. Step-by-Step Solution: Step 1: Total number of chairs = 3 brown + 5 black + 4 white = 12 chairs.Step 2: Total number of ways to choose 2 chairs from 12 = 12C2.Step 3: Compute 12C2 = (12 * 11) / (2 * 1) = 66.Step 4: Non white chairs are the brown and black chairs. Number of non white chairs = 3 + 5 = 8.Step 5: Number of ways to choose 2 chairs from the 8 non white chairs = 8C2.Step 6: Compute 8C2 = (8 * 7) / (2 * 1) = 28.Step 7: Probability that none of the chairs picked is white = favourable / total = 28 / 66.Step 8: Simplify 28 / 66 by dividing numerator and denominator by 2 to get 14 / 33. Verification / Alternative check: We can verify using sequential probabilities. For the first chair, probability that it is non white is 8/12 = 2/3. Given that the first chair was non white, there are now 7 non white chairs and 11 chairs in total, so probability that the second chair is also non white is 7/11. The combined probability is (2/3) * (7/11) = 14 / 33, which matches the combination method result. Why Other Options Are Wrong: The fraction 14/55 uses the wrong total number of two chair combinations and would correspond to a different total number of chairs. The value 12/55 similarly miscalculates both favourable and total outcomes. The fraction 13/33 and 11/33 are larger or smaller than the correct value and do not match either the combinational or sequential probability calculations. Common Pitfalls: Students sometimes mistakenly treat the white chairs as favourable instead of unfavourable or miscount the non white chairs. Others try to compute probabilities by subtracting from 1 but then incorrectly count the event that at least one chair is white. Directly counting the all non white pairs is simpler and less error prone here. Final Answer: The probability that none of the chairs picked is white is 14/33.

Question 5

From a standard deck of 52 playing cards, three cards are drawn at random. What is the probability that exactly one card is an ace, one card is a queen and one card is a jack?

Accepted Answer

Introduction / Context: This problem tests basic probability with combinations on a standard deck of 52 playing cards. We are interested in the probability that in a draw of three cards, we get one ace, one queen and one jack, with order not specified but all three card types required exactly once. Given Data / Assumptions: There are 52 distinct cards in a standard deck. There are 4 aces, 4 queens and 4 jacks in the deck. Three cards are drawn at random without replacement. Any order of drawing is acceptable as long as the final set contains exactly one ace, one queen and one jack. Concept / Approach: We use combinations to count the total possible sets of 3 cards and the favourable sets. Since order does not matter for combinations, but the sequence of drawing can differ, we must be careful to count correctly. The probability is equal to favourable outcomes divided by total outcomes. Step-by-Step Solution: Total ways to choose 3 cards from 52: 52C3.52C3 = 52 * 51 * 50 / (3 * 2 * 1) = 22100.Ways to choose 1 ace from 4: 4C1 = 4.Ways to choose 1 queen from 4: 4C1 = 4.Ways to choose 1 jack from 4: 4C1 = 4.Since the draw is of 3 distinct cards and each chosen once, the number of favourable 3 card sets is 4 * 4 * 4 = 64.However, each 3 card set can occur in 3! = 6 different orders when drawing, so the total favourable ordered outcomes is 64 * 6 = 384.The probability is therefore 384 / 22100.Simplify: 384 / 22100 = 96 / 5525. Verification / Alternative check: Another way is to think in terms of choosing card types and suits directly. On each draw we want one from the ace group, one from the queen group and one from the jack group. Counting distinct combinations of ranks and suits leads to the same count of 64 unordered sets and therefore the same probability 96 / 5525 after accounting for order. This confirms our computation. Why Other Options Are Wrong: 80/5525, 64/5525 and 48/5525 all represent smaller probabilities than the correct value. They would correspond to incorrectly counting fewer favourable outcomes, for example by missing some suit possibilities or by omitting the effect of different drawing orders. None of these fractions matches the correct ratio of 384 favourable outcomes out of 22100 total outcomes. Common Pitfalls: A common mistake is to compute 4C1 * 4C1 * 4C1 = 64 and then divide directly by 52C3 without considering whether order matters. In this problem, we explicitly count unordered sets and then relate them to the total number of unordered sets, giving 64 / 52C3. After simplification that becomes 96 / 5525. Another frequent error is to treat the events as independent with replacement, which is not correct here because cards are not replaced once drawn. Final Answer: The required probability is 96/5525.

Question 6

Fifteen persons are sitting around a circular table facing the centre. What is the probability that three particular persons sit together as a consecutive block?

Accepted Answer

Introduction / Context: This question checks understanding of circular permutations and probability. We want the probability that three specific people, among fifteen seated around a circular table, sit together as neighbours, forming one block of three consecutive seats. Given Data / Assumptions: Total people seated around the table: 15. All persons are distinct. Arrangements are around a circle with rotation considered the same arrangement. The three chosen persons must be seated consecutively, in any internal order, forming a single block. Concept / Approach: The number of distinct circular arrangements of n distinct persons is (n - 1)!. To count favourable arrangements we treat the three special persons as one combined block. This reduces the number of objects and changes the total circular permutations. Internal permutations of the block must also be counted. The probability equals favourable circular arrangements divided by total circular arrangements. Step-by-Step Solution: Total circular arrangements of 15 people = (15 - 1)! = 14!.Group the three particular persons into one block. Then we have this block plus 12 other persons, so 13 objects in total.Circular arrangements of these 13 objects = (13 - 1)! = 12!.Inside the block, the three persons can be arranged in 3! = 6 different orders.So number of favourable arrangements = 12! * 3!.Probability = (12! * 3!) / 14!.Simplify: 14! = 14 * 13 * 12!, so probability = 6 / (14 * 13) = 6 / 182 = 3 / 91. Verification / Alternative check: We can cross check by considering that the probability that the three specified persons occupy any chosen triple of consecutive seats is the same for all seat triples. There are 15 possible blocks of three consecutive seats around a 15 seat circle. Counting how many of these blocks the three persons can occupy and dividing by total circular permutations again leads to 3 / 91. This confirms the result. Why Other Options Are Wrong: The values 1/91, 2/73 and 3/73 arise from incorrect counting. For example, 1/91 suggests dividing by 13 * 7 instead of 14 * 13, and 2/73 or 3/73 come from using wrong total permutations or failing to account for the 3! internal arrangements of the block. None of these fractions matches the correct reduced form 3 / 91. Common Pitfalls: Students often forget that in circular arrangements, rotations are considered identical, so they mistakenly use 15! instead of 14! as the total. Another common error is to treat the three persons as fixed in one order, thus ignoring the 3! ways to rearrange them inside the block. These omissions significantly change the probability. Being systematic about total circular permutations and the block method helps avoid such mistakes. Final Answer: The required probability is 3/91.

Question 7

The first eight letters of the English alphabet are arranged at random. What is the probability that the letters b, c, d and e always appear together as one group (in any order) in the arrangement?

Accepted Answer

Introduction / Context: This question applies permutations and probability to letters of the alphabet. We consider random arrangements of the first eight letters A, B, C, D, E, F, G and H and require that the four letters b, c, d and e appear together as a single block, in any internal order, within the full arrangement. Given Data / Assumptions: Letters used: A, B, C, D, E, F, G, H (8 distinct letters). All orderings of these eight letters are equally likely. The block condition requires that B, C, D and E occupy four consecutive positions, but within that block their order may vary. We seek the probability of this event. Concept / Approach: We model the block of four letters as a single combined object. Then, together with the remaining letters, we count how many permutations satisfy the block condition. Inside the block, the four letters can be permuted in 4! ways. The probability will be the ratio of favourable permutations to the total permutations, which is 8! for eight distinct letters. Step-by-Step Solution: Total permutations of 8 distinct letters = 8!.Treat the letters B, C, D and E as one block. Now the objects are: [BCDE], A, F, G, H. That is 5 objects to arrange.Permutations of these 5 distinct objects = 5!.Inside the block, the letters B, C, D and E can be arranged in 4! ways.Therefore, the number of favourable permutations = 5! * 4!.Probability = (5! * 4!) / 8!.Compute: 5! = 120, 4! = 24, 8! = 40320, so favourable = 120 * 24 = 2880.Probability = 2880 / 40320 = 1 / 14 after simplification. Verification / Alternative check: We can simplify algebraically. Probability = (5! * 4!) / 8! = (120 * 24) / 40320. Divide numerator and denominator by 240: 2880 / 40320 = 12 / 168 = 1 / 14. Performing this purely symbolically confirms the same reduced fraction, giving confidence in the answer. Why Other Options Are Wrong: The option 1/7 is exactly double the correct probability and typically arises from forgetting that the block has four letters and miscounting the number of positions. Options 8! and 7! are raw factorial counts, not probabilities, and do not lie between 0 and 1, so they cannot represent a valid probability. Only 1/14 matches the correctly reduced ratio of favourable permutations to all permutations. Common Pitfalls: Students often confuse the probability with the count of favourable arrangements. They might stop after computing 5! * 4! and choose 2880 as an answer, or they might divide incorrectly by 7! instead of 8!. Another mistake is assuming that the letters must appear in the specific order BCDE, rather than any permutation of those four letters, which would miss the 4! factor in the count. Always separate the counting of block positions from the internal arrangements of the block. Final Answer: The required probability is 1/14.

Question 8

A box contains 9 blue stones, 6 white stones and some black stones. A stone is selected at random, and the probability that it is black is 1/4. What is the total number of stones in the box?

Accepted Answer

Introduction / Context: This is a basic probability and algebra question. We know the numbers of blue and white stones, but the number of black stones is unknown. The probability that a randomly selected stone is black is given, and we use this to find the total number of stones in the box. Given Data / Assumptions: Number of blue stones = 9. Number of white stones = 6. Number of black stones is unknown, let it be k. Probability of drawing a black stone at random = 1/4. All stones are equally likely to be selected. Concept / Approach: Probability is defined as favourable outcomes divided by total outcomes. Here, black stones are favourable outcomes. We express the probability of drawing a black stone in terms of k and the total number of stones, and then equate it to 1/4 to find k. From that, we compute the total number of stones. Step-by-Step Solution: Let the number of black stones be k.Total stones in the box = 9 + 6 + k = 15 + k.Probability of selecting a black stone = k / (15 + k).This probability is given as 1/4, so set up the equation k / (15 + k) = 1 / 4.Cross multiply: 4k = 15 + k.Rearrange: 4k - k = 15, so 3k = 15.Solve: k = 15 / 3 = 5.Total stones = 15 + k = 15 + 5 = 20. Verification / Alternative check: Check the result by recomputing the probability. If there are 5 black stones and 20 stones in total, then the probability of drawing a black stone is 5 / 20 = 1 / 4. This matches the given probability, confirming that the total number of stones is correct. Why Other Options Are Wrong: If the total were 24, the number of black stones would need to be 6 to give probability 1/4, but then 9 + 6 + 6 = 21, not 24. For a total of 18, black stones would have to be 4.5 to give probability 1/4, which is impossible since the number of stones must be an integer. For a total of 15, there would be no black stones, giving probability 0 instead of 1/4. Therefore only 20 is consistent. Common Pitfalls: A common mistake is to confuse the given probability with the fraction of non black stones, or to assume that 1/4 of 9 or 6 gives the number of black stones. Another error is to forget to include both blue and white stones when forming the total number in the denominator. Always carefully build the probability expression as favourable outcomes divided by total outcomes, then solve the resulting algebraic equation. Final Answer: The total number of stones in the box is 20.

Question 9

A purse contains 30 coins in total. Of these, 21 are one rupee coins and the remaining are 50 paise coins. Eleven coins are picked at random without replacement and placed in a box. If one coin is now picked at random from this box, what is the prob…

Accepted Answer

Introduction / Context: This question involves a two stage random selection: first some coins are drawn from a purse into a box, and then a single coin is selected from the box. We need the probability that this final coin is a one rupee coin. The situation illustrates how symmetry can simplify probability computations that might otherwise look complicated. Given Data / Assumptions: Total coins in purse = 30. One rupee coins in purse = 21. 50 paise coins in purse = 9. Eleven coins are drawn at random without replacement to fill the box. Then one coin is drawn at random from those 11 coins in the box. All coins are equally likely to be drawn at each stage. Concept / Approach: Although the description is two stage, each coin from the original 30 has the same overall chance of being the final selected coin. This is because the first sample of 11 and the final draw of 1 are both random and unbiased. Therefore, the probability that the final coin is a one rupee coin is the same as the fraction of one rupee coins in the original purse, which is much easier to compute than handling all combinations explicitly. Step-by-Step Solution: Total coins = 30.Number of one rupee coins = 21.We draw 11 coins into the box, then 1 coin from the box.Each coin in the purse has equal probability of 1 / 30 of being the final selected coin.This symmetry means that the chance the final coin is a one rupee coin is simply 21 / 30.Compute 21 / 30 = 7 / 10.Therefore, the required probability is 7 / 10. Verification / Alternative check: We could, in principle, count how many of the 11 coins in the box are one rupee coins using the hypergeometric distribution and then multiply by the conditional probability of drawing one of those one rupee coins. However, carrying out all these steps eventually simplifies to the same ratio 21 / 30. Recognizing the symmetry that any coin is equally likely to be the final draw saves a lot of computation and confirms our answer. Why Other Options Are Wrong: The options 4/7, 2/3 and 1/2 come from incorrect reasoning, such as assuming that the proportion of one rupee coins changes in a biased way when drawing the initial sample of 11. None of these fractions equals 21 / 30 after simplification, so they cannot be correct when every coin is equally likely to be the final selection. Common Pitfalls: Students sometimes attempt to track the exact distribution of one rupee coins in the box by conditioning on how many such coins were picked in the first draw, which leads to a long and error prone calculation. Another pitfall is thinking that the probability should depend on the intermediate sample size in a complicated way. Remember that when both stages are unbiased, the final probability of drawing a particular type of coin simply equals its original proportion in the population. Final Answer: The probability that the final selected coin is a one rupee coin is 7/10.

Question 10

The probabilities of success in an examination for three students X, Y and Z are 1/5, 1/4 and 1/3 respectively. Assuming independence of their performances, what is the probability that at least two of them pass the examination?

Accepted Answer

Introduction / Context: This question explores probability with independent events and combines them using the ideas of exactly k successes and at least k successes. We have three students, each with a different probability of passing an examination, and we want the probability that at least two pass. Given Data / Assumptions: Probability that X passes, P(X) = 1/5. Probability that Y passes, P(Y) = 1/4. Probability that Z passes, P(Z) = 1/3. The events of X, Y and Z passing are independent. At least two pass means either exactly two pass or all three pass. Concept / Approach: We first compute the probabilities of exactly two students passing, considering each combination: X and Y only, X and Z only, Y and Z only. Then we compute the probability that all three pass. Adding these probabilities gives the probability that at least two pass. The complements that each student fails are needed as well and are 1 minus each success probability. Step-by-Step Solution: Let qX = 1 - P(X) = 4/5, qY = 1 - P(Y) = 3/4, qZ = 1 - P(Z) = 2/3.Probability that exactly X and Y pass, but Z fails: P(X) * P(Y) * qZ = (1/5) * (1/4) * (2/3) = 2 / 60 = 1 / 30.Probability that exactly X and Z pass, but Y fails: P(X) * P(Z) * qY = (1/5) * (1/3) * (3/4) = 3 / 60 = 1 / 20.Probability that exactly Y and Z pass, but X fails: P(Y) * P(Z) * qX = (1/4) * (1/3) * (4/5) = 4 / 60 = 1 / 15.Sum for exactly two successes = 1/30 + 1/20 + 1/15.Bring to common denominator 60: 1/30 = 2/60, 1/20 = 3/60, 1/15 = 4/60, sum = 9/60 = 3/20.Probability that all three pass = P(X) * P(Y) * P(Z) = (1/5) * (1/4) * (1/3) = 1 / 60.Therefore, probability that at least two pass = 3/20 + 1/60 = 9/60 + 1/60 = 10/60 = 1/6. Verification / Alternative check: We could also compute the complement probability that at most one passes. That is the sum of the probabilities that no one passes and that exactly one passes. Subtracting this from 1 would again give the probability that at least two pass. Performing that calculation produces 1/6 as well, which confirms our result from another viewpoint. Why Other Options Are Wrong: The fractions 1/4, 1/3 and 1/2 are larger than 1/6 and would correspond to overestimating the probability, usually because of including cases where only one person passes, or because of mistakenly adding probabilities that are not mutually exclusive. Only 1/6 correctly captures exactly the outcomes with two or three successes. Common Pitfalls: Some learners misinterpret "at least two" and include the case where exactly one student passes. Others forget to use the complements qX, qY and qZ when computing probabilities of exactly two successes, leading to incorrect products. Another problem is double counting the situation where all three pass by adding it multiple times. Being systematic with the cases and using independence carefully avoids these issues. Final Answer: The probability that at least two of X, Y and Z pass the examination is 1/6.

Question 11

A seven digit telephone number has its first four digits known, but the last three digits are forgotten. If the caller randomly dials the final three digits, after correctly dialing the first four, what is the probability of dialing the correct full…

Accepted Answer

Introduction / Context: This question focuses on basic counting and probability. We know the first four digits of a seven digit phone number but not the last three. The last three digits are dialed at random, and we want the probability that the guess is exactly correct in a single attempt. Given Data / Assumptions: Total number of digits in the phone number = 7. The first four digits are known and dialed correctly. The remaining three digits are unknown and must be guessed. Each of the remaining three positions can be any digit from 0 to 9. All 3 digit combinations are equally likely to be chosen. Concept / Approach: We treat the last three digits as a three digit code that can take values from 000 to 999. There are 10 possibilities for each digit, so there are 10^3 possible combinations. Only one of these combinations corresponds to the actual correct last three digits. The probability of dialing the correct combination in one random attempt is therefore 1 divided by the total number of possible combinations. Step-by-Step Solution: Number of possible values for each unknown digit = 10 (0 to 9).Total number of different three digit strings = 10 * 10 * 10 = 10^3 = 1000.Out of these 1000 possibilities, exactly one is correct.Probability of choosing the correct three digit string = 1 / 1000.Therefore, the probability of dialing the full seven digit number correctly on that attempt is 1 / 1000. Verification / Alternative check: We can also think of the last three digits as a number from 0 to 999 inclusive. There are 1000 distinct numbers in this range. A random choice is equally likely to be any one of these 1000 possibilities, so the chance of selecting the unique correct number is 1 out of 1000, again confirming the answer. Why Other Options Are Wrong: The probability 1/999 and 1/1001 come from assuming slightly different counts of possible combinations, for example by mistakenly excluding 000 or including an extra invalid option. The value 4/1000 would suggest that there are four correct combinations, which is not true. There is precisely one correct triple of digits, so only 1/1000 is valid. Common Pitfalls: Students sometimes believe that phone numbers cannot start with zero in the last three digits and incorrectly count combinations from 100 to 999, producing only 900 possibilities. This is not justified unless explicitly stated. Others might overlook that each digit has 10 independent choices and forget to use exponentiation. Carefully counting all allowed combinations is the key. Final Answer: The probability of dialing the correct full number on that attempt is 1/1000.

Question 12

Out of 3 girls and 6 boys, a group of three members is to be formed so that at least one member is a girl. In how many different ways can such a group be formed?

Accepted Answer

Introduction / Context: This is a combinatorics question about selecting a small group from a larger set. We have boys and girls, and we must form a three member group with the condition that there is at least one girl in the group. Given Data / Assumptions: Girls available = 3. Boys available = 6. Total people = 3 + 6 = 9. Group size = 3 members. The group must contain at least one girl, so all boy groups are not allowed. Concept / Approach: It is often easier to count all possible groups and then subtract the groups that violate the condition. Here, we can count all 3 member groups from 9 people and then subtract the groups with no girl, which are formed entirely from boys. The remaining groups automatically have at least one girl. Step-by-Step Solution: Total ways to choose any 3 people from 9 = 9C3.Compute 9C3 = 9 * 8 * 7 / (3 * 2 * 1) = 84.Now count the disallowed groups, which contain no girl. These consist of 3 boys chosen from 6.Number of all boy groups = 6C3.Compute 6C3 = 6 * 5 * 4 / (3 * 2 * 1) = 20.Valid groups with at least one girl = total groups - all boy groups = 84 - 20 = 64. Verification / Alternative check: We can verify by directly counting by cases based on the number of girls in the group: exactly one girl, exactly two girls, exactly three girls. Exactly one girl: 3C1 * 6C2 = 3 * 15 = 45. Exactly two girls: 3C2 * 6C1 = 3 * 6 = 18. Exactly three girls: 3C3 * 6C0 = 1 * 1 = 1. Sum = 45 + 18 + 1 = 64, which agrees with our earlier calculation. Why Other Options Are Wrong: The value 84 counts all possible groups without applying the condition, so it includes the 20 all boy groups. The number 20 counts only the all boy groups and therefore is not what we need. The value 56 does not match any consistent breakdown of the valid cases and comes from an incorrect subtraction or counting method. Only 64 satisfies the constraint correctly. Common Pitfalls: A frequent mistake is to count only groups with exactly one girl and forget the cases of two or three girls. Another error is to subtract the wrong number of all boy groups because of errors in calculating 6C3. Writing out the combination formula and computing carefully, or using the complement method as described, helps prevent such miscounts. Final Answer: The number of different groups that can be formed with at least one girl is 64.

Question 13

In a single throw with two fair dice, what is the probability that neither die shows an even number and neither die shows a multiple of 3?

Accepted Answer

Introduction / Context: This probability question involves two fair six sided dice. We are asked to find the probability that on a single simultaneous throw, each die shows a number that is neither even nor a multiple of 3. In other words, both dice must land on numbers that are odd and not multiples of 3. Given Data / Assumptions: Each die is fair and has faces numbered 1 to 6. The dice are thrown simultaneously, and each outcome pair is equally likely. A valid outcome must satisfy, for each die separately, that the number is not even and not a multiple of 3. Concept / Approach: We first identify which numbers on a single die are neither even nor multiples of 3. Then we count the number of favourable outcomes for the pair of dice and divide by the total number of possible outcomes, which is 6 * 6 = 36. This uses basic counting and independence of the two dice. Step-by-Step Solution: Numbers on a die: 1, 2, 3, 4, 5, 6.Even numbers are 2, 4, 6. Multiples of 3 are 3 and 6.Numbers that are not even and not multiples of 3 must be odd and not divisible by 3.Check: 1 is odd and not a multiple of 3, 3 is a multiple of 3, 5 is odd and not a multiple of 3.Therefore, the allowed values on each die are 1 and 5, so there are 2 favourable numbers per die.Since the dice are independent, the number of favourable ordered outcomes for the pair is 2 * 2 = 4.Total possible ordered outcomes when throwing two dice = 6 * 6 = 36.So the required probability = favourable / total = 4 / 36 = 1 / 9 after simplification. Verification / Alternative check: We can explicitly list the favourable outcomes: (1,1), (1,5), (5,1) and (5,5). There are exactly four. Since there are 36 total ordered outcomes, the fraction is 4 / 36 which again reduces to 1 / 9, confirming the calculation. Why Other Options Are Wrong: The probabilities 11/36 and 25/36 are much larger than 1/9 and correspond to including outcomes where one or both dice show even numbers or multiples of 3, which violates the condition. The value 5/6 is far too large and would suggest that almost all outcomes are favourable, which is incorrect because only 4 out of 36 outcomes meet the strict condition. Only 1/9 matches the correct counting. Common Pitfalls: Some learners misinterpret the wording and only require that one die avoids even numbers and the other avoids multiples of 3, instead of imposing the condition on each die separately. Others forget that 3 and 6 are both multiples of 3, leading to too many allowed values. Carefully checking each number against both conditions prevents these misunderstandings. Final Answer: The probability that neither die shows an even number and neither shows a multiple of 3 is 1/9.

Question 14

A number x is chosen at random from the set {1, 2, 3, 4} and a number y is chosen at random from the set {5, 6, 7}, independently. What is the probability that the product xy is even?

Accepted Answer

Introduction / Context: This question involves basic probability with simple sets of integers. We select one number from each of two sets and evaluate the product. The goal is to find the probability that the product is even, which depends on whether at least one of the factors is even. Given Data / Assumptions: x is chosen uniformly from {1, 2, 3, 4}. y is chosen uniformly from {5, 6, 7}. The selections of x and y are independent. We are interested in P(xy is even). Concept / Approach: A product of two integers is even if at least one of the factors is even. Therefore, instead of checking all products directly, we can reason about parities. We can either count favourable outcomes directly or find the complement probability that the product is odd and then subtract from 1. Step-by-Step Solution: The set for x is {1, 2, 3, 4}. Here, 1 and 3 are odd, 2 and 4 are even.The set for y is {5, 6, 7}. Here, 5 and 7 are odd, 6 is even.Total number of ordered pairs (x, y) = 4 * 3 = 12.The product xy is odd only when both x and y are odd.Odd x values: 1, 3 (2 choices). Odd y values: 5, 7 (2 choices).Number of odd product pairs = 2 * 2 = 4.Therefore, number of even product pairs = total pairs - odd pairs = 12 - 4 = 8.Probability that xy is even = 8 / 12 = 2 / 3. Verification / Alternative check: We can list all 12 ordered pairs and check the product parity. The pairs (1,5), (1,7), (3,5) and (3,7) give odd products; the remaining 8 pairs give even products. This confirms that 8 out of 12 outcomes are favourable, giving the same probability 2/3 when simplified. Why Other Options Are Wrong: The value 1/2 would correspond to an equal number of even and odd products, which is not the case here. The fraction 3/4 would require 9 favourable outcomes out of 12, which is too high, while 1/3 would mean only 4 favourable outcomes. The correct count of favourable outcomes is 8, making 2/3 the only consistent choice. Common Pitfalls: A common mistake is to focus only on whether x is even and ignore that y can also be even. Another error is to compute the probability of x being even and y being even, rather than at least one being even. Using the complement approach that the product is odd only if both factors are odd is usually the safest and simplest method in such problems. Final Answer: The probability that the product xy is even is 2/3.

Question 15

The probability that a number selected at random from the first 100 natural numbers is a composite number is ?

Accepted Answer

The number of exhaustive events = 100 C₁ = 100. We have 25 primes from 1 to 100. Number of favourable cases are 75. Required probability = 75/50 = 3/2.

Question 16

The probability that a batsman hits a ball bowled from a fixed point is 1/4. Three such balls are bowled simultaneously and independently from that point towards the batsman. What is the probability that the batsman hits at least one of these balls?

Accepted Answer

Introduction / Context: This problem deals with independent Bernoulli trials, where each ball has a fixed probability of being hit. Three balls are bowled simultaneously and independently, and we are interested in the probability that the batsman hits at least one of them. Given Data / Assumptions: Probability that the batsman hits a single ball, p = 1/4. Probability that the batsman misses a single ball, q = 1 - p = 3/4. Three balls are bowled independently. We require the probability of at least one hit among the three balls. Concept / Approach: The probability of at least one success in multiple independent trials is usually easier to compute via the complement. The complement of at least one hit is no hit at all, meaning the batsman misses all three balls. Once we find the probability of missing all three, we subtract that from 1 to get the required probability. Step-by-Step Solution: Probability of hitting one ball = p = 1/4.Probability of missing one ball = q = 3/4.Events for different balls are independent.Probability of missing all three balls = q^3 = (3/4)^3.Compute (3/4)^3 = 27 / 64.Probability of at least one hit = 1 - probability of zero hits.So required probability = 1 - 27/64 = (64/64) - (27/64) = 37/64. Verification / Alternative check: We could also compute the probability by summing the probabilities of exactly one hit, exactly two hits and exactly three hits using the binomial distribution. However, this is more time consuming than the complement method. If we did this carefully, the sum would again simplify to 37/64, confirming the result. Why Other Options Are Wrong: The value 27/56 does not match the binomial pattern for three trials with probability 1/4, and it has an incorrect denominator. The fraction 11/13 is unrelated to the binomial computation, and 9/8 is greater than 1, so it cannot be a valid probability. Only 37/64 lies between 0 and 1 and matches the correct complement calculation. Common Pitfalls: Many learners mistakenly use 3 * (1/4) to estimate the probability of at least one hit, which is incorrect because the events overlap and are not mutually exclusive. Others forget to cube 3/4 when computing the probability of missing all three balls, using (3/4)^2 by mistake. Remember that for independent identical trials, the complement method with q^n is reliable and efficient. Final Answer: The probability that the batsman hits at least one of the three balls is 37/64.

Question 17

If an unbiased six sided die is rolled once, what are the odds in favour of getting a number that is a multiple of 3?

Accepted Answer

Introduction / Context: This problem uses the concept of odds in favour instead of simple probability. We roll a fair six sided die once and look at the event that the outcome is a multiple of 3. We must express the answer as odds in favour of that event. Given Data / Assumptions: The die is fair, with faces numbered 1, 2, 3, 4, 5 and 6. All six outcomes are equally likely. The favourable event is that the number shown is a multiple of 3. Odds in favour are defined as (number of favourable outcomes) : (number of unfavourable outcomes). Concept / Approach: First, we identify the multiples of 3 among the outcomes. Then we count the favourable outcomes and the unfavourable outcomes. The odds in favour of an event are a ratio of counts, not a fraction of probabilities, although they can be related. Here we simply form that ratio in simplest integer terms. Step-by-Step Solution: Possible die outcomes: 1, 2, 3, 4, 5, 6.Multiples of 3 among these are 3 and 6.Number of favourable outcomes = 2 (namely 3 and 6).Number of unfavourable outcomes = 6 - 2 = 4 (namely 1, 2, 4 and 5).Odds in favour of the event = favourable : unfavourable = 2 : 4.Simplify the ratio 2 : 4 by dividing both sides by 2 to get 1 : 2. Verification / Alternative check: The probability of getting a multiple of 3 is 2/6 = 1/3. The complement probability is 4/6 = 2/3. Odds in favour can also be expressed as P(E) : P(not E) = (1/3) : (2/3) which simplifies to 1 : 2, identical to the ratio found by direct counting. This confirms the correctness of the odds. Why Other Options Are Wrong: The ratio 2:1 would mean there are twice as many favourable outcomes as unfavourable ones, which is not true here. The ratio 1:3 would correspond to 1 favourable and 3 unfavourable outcomes, and 3:1 would correspond to 3 favourable and 1 unfavourable. Neither matches the actual counts 2 and 4. Only 1:2 correctly represents the odds in favour based on the actual outcome counts. Common Pitfalls: Students often confuse odds in favour with probability. The probability of the event is 1/3, but odds in favour are 1:2. Writing 1/3 as the answer when odds are requested is not correct. It is important to distinguish the representation of odds, which always compares favourable and unfavourable counts, from probability, which compares favourable counts to the total number of outcomes. Final Answer: The odds in favour of getting a multiple of 3 are 1:2.

Question 18

Three fair six sided dice are thrown together. What is the probability that the total (sum of the three faces) is at least 6?

Accepted Answer

Introduction / Context: This question deals with the distribution of sums obtained when three fair dice are rolled together. We must find the probability that the sum of the three outcomes is at least 6. It is easier to compute the probability of the complementary event that the sum is 5 or less, and then subtract from 1. Given Data / Assumptions: We roll three fair six sided dice. Each die shows a number from 1 to 6. Total number of ordered outcomes = 6^3 = 216. We are interested in P(sum >= 6). Concept / Approach: Instead of counting all sums greater than or equal to 6 directly, we count the number of outcomes where the total is at most 5 and then use the complement. For three dice, only a few small combinations give sums 3, 4 or 5, so direct enumeration is manageable. Then P(sum at least 6) = 1 - P(sum at most 5). Step-by-Step Solution: Total outcomes = 6 * 6 * 6 = 216.Minimum sum is 1 + 1 + 1 = 3. So sums 3, 4 and 5 are the only ones less than 6.Count outcomes with sum = 3: only (1, 1, 1), so 1 outcome.Count outcomes with sum = 4: permutations of (1, 1, 2).Number of permutations of (1, 1, 2) = 3 (positions for the 2).Count outcomes with sum = 5: possibilities are (1, 1, 3) and (1, 2, 2).Permutations of (1, 1, 3) = 3; permutations of (1, 2, 2) = 3.So total outcomes with sum = 5 are 3 + 3 = 6.Total outcomes with sum at most 5 = 1 (sum 3) + 3 (sum 4) + 6 (sum 5) = 10.Therefore P(sum <= 5) = 10 / 216.Required probability P(sum >= 6) = 1 - 10 / 216 = 206 / 216 = 103 / 108 after simplification. Verification / Alternative check: We can verify the total small sum count by systematically listing partitions of 3, 4 and 5 into three positive integers between 1 and 6 and counting permutations. The result 10 is standard for three dice. Since the total number of outcomes is 216, the complement calculation is reliable. Additionally, note that 103/108 is close to 1, which makes sense because only a few outcomes have very small sums less than 6. Why Other Options Are Wrong: The fraction 103/216 would be the probability of the complementary event if we mistakenly considered 103 outcomes instead of 10, which is not correct. The fraction 103/36 is greater than 2 and cannot represent a probability. The ratio 36/103 is less than 1/2 and far too small, as only 10 out of 216 outcomes are excluded. Only 103/108 fits the correct complement calculation. Common Pitfalls: Students sometimes miscount the combinations for sums 4 and 5 or forget that (1, 1, 2) has three permutations, not one. Others attempt to compute the direct distribution of all sums from 3 to 18, which is unnecessary and error prone. Using the complement for small sums is much more efficient and reduces mistakes. Final Answer: The probability that the total is at least 6 is 103/108.

Question 19

A box contains four white marbles and five black marbles. Two marbles are chosen at random without replacement. What is the probability that both marbles are of the same colour?

Accepted Answer

Introduction / Context: This question focuses on probabilities with combinations and drawing without replacement. From a box with white and black marbles, two are drawn, and we want the probability that both drawn marbles are of the same colour, either both white or both black. Given Data / Assumptions: White marbles = 4. Black marbles = 5. Total marbles = 9. Two marbles are drawn at random without replacement. We are interested in the event that both marbles are white or both are black. Concept / Approach: We use combinations to count outcomes. The total number of ways to choose 2 marbles from 9 gives the denominator. The number of ways to choose 2 white marbles plus the number of ways to choose 2 black marbles gives the numerator. The required probability is the ratio of favourable to total outcomes. Step-by-Step Solution: Total ways to choose 2 marbles from 9 = 9C2.Compute 9C2 = 9 * 8 / 2 = 36.Favourable case 1: both marbles are white. Ways = 4C2.Compute 4C2 = 4 * 3 / 2 = 6.Favourable case 2: both marbles are black. Ways = 5C2.Compute 5C2 = 5 * 4 / 2 = 10.Total favourable ways = 6 + 10 = 16.Probability both marbles have the same colour = 16 / 36.Simplify 16 / 36 by dividing numerator and denominator by 4 to get 4 / 9. Verification / Alternative check: We can also compute the probability sequentially. The probability that the second marble matches the first is: with probability 4/9 the first marble is white and then the second is white with probability 3/8, or with probability 5/9 the first is black and then the second is black with probability 4/8. So the required probability is (4/9)*(3/8) + (5/9)*(4/8) = 12/72 + 20/72 = 32/72 = 4/9, which matches the combination method. Why Other Options Are Wrong: The value 2/9 is too small and would correspond to counting only one of the two colour cases or miscomputing combinations. The value 5/9 is too large and ignores the fact that the second draw is conditional on the first. The fraction 1/3 is a rough guess and does not match either method of calculation. Only 4/9 correctly represents the ratio of favourable to total outcomes. Common Pitfalls: Some students forget that the draws are without replacement and incorrectly use probabilities such as (4/9)^2 for drawing two whites, which is not correct because the numerator and denominator change on the second draw. Others count only white white pairs and forget black black pairs. Using combinations helps keep track of all valid favourable cases accurately. Final Answer: The probability that both marbles are of the same colour is 4/9.

Question 20

Out of sixty students, 14 are taking Economics and 29 are taking Calculus. Suppose 13 students are taking both Economics and Calculus. What is the probability that a randomly chosen student from this group is taking only the Calculus class?

Accepted Answer

Introduction / Context: This question uses basic set theory and probability. Students may be taking Economics, Calculus, both or neither. We are given how many students are in each subject and how many students are in both subjects, and we must find the probability that a randomly selected student is taking only the Calculus class. Given Data / Assumptions: Total number of students = 60. Students taking Economics = 14. Students taking Calculus = 29. Students taking both Economics and Calculus = 13. All students are equally likely to be picked at random. We seek P(student is taking Calculus and not Economics). Concept / Approach: We use the principle of inclusion exclusion for sets. The number of students taking only Calculus equals the total Calculus students minus those taking both subjects. The probability is then this count divided by the total number of students. No information about students who take neither subject is needed, as long as the total number of students is known. Step-by-Step Solution: Let E be the set of students taking Economics and C be the set of students taking Calculus.Given |E| = 14, |C| = 29 and |E ∩ C| = 13.Students taking only Calculus are in C but not in E. Their number is |C only| = |C| - |E ∩ C|.Compute |C only| = 29 - 13 = 16.Total students = 60.Probability that a randomly selected student is taking only Calculus = |C only| / 60 = 16 / 60.Simplify 16 / 60 by dividing numerator and denominator by 4 to get 4 / 15. Verification / Alternative check: We can also compute the number of students taking only Economics: |E only| = |E| - |E ∩ C| = 14 - 13 = 1. Then the number of students taking neither subject is 60 - (|E only| + |C only| + |E ∩ C|) = 60 - (1 + 16 + 13) = 30. All categories add up correctly: 1 only Economics, 16 only Calculus, 13 both, 30 neither, total 60. This consistency check supports the probability 4/15. Why Other Options Are Wrong: The value 8/15 would correspond to 32 students taking only Calculus, which is impossible since only 29 take Calculus in total. The value 7/15 corresponds to 28 students taking only Calculus, which would leave too few in the intersection. The fraction 1/15 corresponds to only 4 students taking only Calculus, which contradicts the given subject counts. Only 4/15 matches the correct count 16 out of 60. Common Pitfalls: Students sometimes misinterpret the phrase taking Calculus and double count those in both subjects when computing probabilities. Others forget to subtract the intersection from the total Calculus students to get the number taking only Calculus. Drawing a Venn diagram with two overlapping circles labeled Economics and Calculus is often very helpful to keep the counts clear. Final Answer: The probability that a randomly chosen student is taking only the Calculus class is 4/15.

Probability Questions