Question 1

One card is drawn at random from a standard pack of 52 playing cards, each card being equally likely. What is the probability that the card drawn is an ace?

Accepted Answer

Introduction / Context: This is one of the simplest and most common probability questions asked with playing cards. It focuses on finding the probability that a single card drawn at random from a full deck is an ace. The key is to know how many aces there are in the deck and then use the basic probability formula based on equally likely outcomes. Given Data / Assumptions: A standard deck contains 52 cards in total. There are 4 suits: hearts, diamonds, clubs and spades. Each suit contains exactly one ace, so there are 4 aces in the deck. One card is drawn at random, and all 52 cards are equally likely to be drawn. We want the probability that the selected card is an ace. Concept / Approach: In a uniform sample space where each card has equal probability, the probability of drawing a card with a specific property is equal to the number of cards with that property divided by the total number of cards. Here the property is being an ace, and we know there are four such cards in the deck. Step-by-Step Solution: Number of aces in a standard deck = 4. Total number of cards = 52. Required probability of drawing an ace = favourable outcomes / total outcomes. So the probability is 4 / 52. Simplify 4 / 52 by dividing numerator and denominator by 4 to obtain 1 / 13. Verification / Alternative check: We can think of this as choosing a card by first choosing a rank and then a suit. There are 13 ranks in the deck, and exactly one of them is the ace rank. So the chance that the chosen card has rank ace is 1 out of 13, which directly gives probability 1 / 13. This reasoning matches our earlier fraction and confirms that the answer is correct. Why Other Options Are Wrong: 1/12 would correspond to about 4.33 aces in the deck, which is not possible. 1/14 and 1/15 are even smaller and suggest fewer than 4 favourable cards, which again contradicts the structure of a standard deck. Common Pitfalls: A typical mistake is confusing aces with face cards and assuming there are more or fewer aces than there really are. Another occasional error is to mix up the total number of cards, for example using 54 when thinking about decks with jokers, even though jokers are not part of a standard 52 card deck in most probability questions. Clarifying deck composition before starting the calculation helps avoid these problems. Final Answer: The probability that the card drawn is an ace is 1/13.

Question 2

The probabilities that persons A and B will tell the truth about an event are 2/3 and 4/5 respectively. Assuming their statements are independent, what is the probability that they will agree with each other?

Accepted Answer

Introduction / Context: This question examines agreement between two people who may or may not tell the truth. Each person has a different probability of telling the truth, and their statements about the same event are assumed to be independent. They are said to agree when both tell the truth or both tell lies. The problem therefore involves calculating the probability of these two joint cases and adding them. Given Data / Assumptions: Probability that A tells the truth = 2/3. Probability that A tells a lie = 1 - 2/3 = 1/3. Probability that B tells the truth = 4/5. Probability that B tells a lie = 1 - 4/5 = 1/5. The behaviour of A and B is independent. They agree if both statements are true or both are false. Concept / Approach: Let T denote telling the truth and L denote telling a lie. Agreement occurs in exactly two cases: A is T and B is T, or A is L and B is L. Since these cases are mutually exclusive, the total probability of agreement is the sum of their individual probabilities. Because A and B act independently, we can find each joint probability by multiplying the corresponding individual probabilities. Step-by-Step Solution: P(A tells truth) = 2/3, P(A tells lie) = 1/3. P(B tells truth) = 4/5, P(B tells lie) = 1/5. Case 1: Both tell truth. P(both truth) = (2/3) * (4/5) = 8 / 15. Case 2: Both tell lies. P(both lies) = (1/3) * (1/5) = 1 / 15. Total probability of agreement = 8 / 15 + 1 / 15 = 9 / 15. Simplify 9 / 15 by dividing numerator and denominator by 3 to get 3 / 5. Verification / Alternative check: We can use the complement idea by first finding the probability that they contradict each other. Contradiction occurs if A tells truth and B lies, or A lies and B tells truth. P(contradiction) = (2/3)*(1/5) + (1/3)*(4/5) = 2/15 + 4/15 = 6/15 = 2/5. Since agreement and contradiction cover all possibilities, P(agreement) = 1 minus 2/5 = 3/5, confirming the earlier result. Why Other Options Are Wrong: 1/3 and 2/5 do not match the combined probability of the two agreement cases and instead relate to partial parts of the outcome space. 3/4 suggests 75 percent agreement, which is too high given that both individuals have a significant chance of lying. Common Pitfalls: Learners sometimes assume agreement means both tell the truth only, forgetting that both can also lie and still give matching statements. Another common error is to add the individual truth probabilities directly rather than multiplying to get joint probabilities. Drawing a small table of outcomes TT, TL, LT and LL helps to visualize when agreement happens and ensures the correct cases are counted. Final Answer: The probability that A and B agree with each other is 3/5.

Question 3

Tickets numbered from 1 to 20 are mixed thoroughly and then one ticket is drawn at random. What is the probability that the number on the drawn ticket is a multiple of 3 or a multiple of 5?

Accepted Answer

Introduction / Context: This question uses basic probability with divisibility rules for integers. The tickets are numbered from 1 to 20, and we want the probability that a randomly selected ticket has a number divisible by 3 or divisible by 5. The main task is to count how many numbers meet this condition and compare that count with the total number of tickets. Given Data / Assumptions: Tickets are numbered 1, 2, 3, up to 20. One ticket is drawn at random, and each ticket has equal chance of selection. Favourable tickets are those with numbers that are multiples of 3 or multiples of 5. Concept / Approach: We apply the union principle for counting. First count the numbers that are multiples of 3, then count those that are multiples of 5, and finally subtract the numbers that are multiples of both 3 and 5, because those have been counted twice. The probability is the ratio of this union count to the total number of tickets, which is 20. Step-by-Step Solution: Total tickets = 20. Multiples of 3 between 1 and 20: 3, 6, 9, 12, 15, 18, which is 6 numbers. Multiples of 5 between 1 and 20: 5, 10, 15, 20, which is 4 numbers. Multiples of both 3 and 5 are multiples of 15: 15 only, which is 1 number. Total favourable numbers = 6 + 4 - 1 = 9. Required probability = 9 / 20. Verification / Alternative check: List all numbers from 1 to 20 and mark those divisible by 3 or 5. The marked numbers are 3, 5, 6, 9, 10, 12, 15, 18 and 20. Counting them gives 9 favourable tickets. Since there are 20 tickets in all, the probability is 9 / 20. The fraction is already in simplest form, so this confirms our earlier counting and ensures there is no arithmetic mistake. Why Other Options Are Wrong: 1/2 would require 10 favourable numbers out of 20, but only 9 numbers meet the condition. 3/5 corresponds to 12 favourable numbers, which is more than we actually have. 2/5 equals 8 favourable numbers, which is slightly less than the correct count of 9. Common Pitfalls: Some learners forget to subtract the overlap of numbers that are multiples of both 3 and 5, leading to 10 instead of 9 favourable numbers. Others may miscount by skipping a multiple or mistakenly including a non multiple. When the range is small, writing out all numbers and checking divisibility carefully is a very effective strategy to avoid these errors. Final Answer: The probability that the drawn ticket shows a number that is a multiple of 3 or 5 is 9/20.

Question 4

Three fair coins are tossed simultaneously. What is the probability of getting more heads than tails in the outcome?

Accepted Answer

Introduction / Context: This probability question deals with tossing three fair coins at the same time. The requirement more heads than tails means that the number of heads in the result must be strictly greater than the number of tails. With three coins, this condition leads to a small set of favourable outcomes, which we can identify by listing the sample space systematically. Given Data / Assumptions: There are three unbiased coins. Each coin has two possible results: head (H) or tail (T). All 2^3 = 8 possible outcomes of the three tosses are equally likely. More heads than tails means heads count greater than tails count. Concept / Approach: For three coins, the number of heads can be 0, 1, 2 or 3. More heads than tails requires heads count greater than tails count. If there are 0 heads, tails are 3, so this is not allowed. If there is 1 head, then tails equal 2, still not allowed. If there are 2 heads, tails are 1 and the condition is satisfied. If there are 3 heads, tails are 0 and the condition is also satisfied. We therefore count outcomes with either 2 heads or 3 heads and divide by the total of 8 outcomes. Step-by-Step Solution: List all outcomes: HHH, HHT, HTH, THH, HTT, THT, TTH, TTT. Outcomes with 2 heads and 1 tail: HHT, HTH, THH, which gives 3 outcomes. Outcome with 3 heads and 0 tails: HHH, which gives 1 outcome. Total favourable outcomes where heads are more than tails = 3 + 1 = 4. Total possible outcomes = 8. Required probability = 4 / 8 = 1 / 2. Verification / Alternative check: We can also use symmetry. For three coins, the distribution of heads is symmetric around 1.5. Exactly half of the outcomes have more heads than tails and half have more tails than or equal number of heads when carefully grouped. Counting confirms that 4 out of 8 outcomes meet the condition, which supports the conclusion that the probability is 1 / 2. Why Other Options Are Wrong: 5/8 would require 5 favourable outcomes, which is more than the 4 that satisfy the condition. 7/8 suggests that almost every outcome has more heads than tails, which is clearly not true when we list them. 3/8 would mean only 3 favourable cases, while we identified 4 outcomes that work. Common Pitfalls: Learners sometimes misinterpret more heads than tails as at least two heads, then forget to include the case of three heads or miscount the number of sequences with two heads. Another frequent mistake is not listing the full sample space and missing one or more outcomes. Writing down all eight possibilities for three coins is quick and ensures complete coverage before counting the favourable ones. Final Answer: The probability of getting more heads than tails when three fair coins are tossed is 1/2.

Question 5

A number is selected at random from the integers 1 to 25 (inclusive). What is the probability that the chosen number is divisible by 4 or by 7?

Accepted Answer

Introduction / Context: Probability questions that involve divisibility over a finite set of integers are standard problems in quantitative aptitude exams. In this question, a number is picked at random from 1 to 25, and we are asked to find the probability that it is divisible by 4 or by 7. This tests understanding of basic counting, divisibility rules, and the idea of favourable outcomes over total outcomes. Given Data / Assumptions: The integers under consideration are 1, 2, 3, ..., 25 (inclusive). Each integer from 1 to 25 is equally likely to be selected. We want numbers divisible by 4 or by 7. Standard probability formula is used: probability = favourable outcomes / total outcomes. Concept / Approach: We first count the total number of integers from 1 to 25. Then we count how many of those integers are divisible by 4, how many are divisible by 7, and check whether any number is divisible by both 4 and 7. The probability is then the number of distinct favourable integers divided by 25. Since 4 and 7 are relatively prime, we only need to check whether any common multiple like 28 falls inside the range, which it does not, so there is no overlap to subtract. Step-by-Step Solution: Total integers from 1 to 25 inclusive = 25. Multiples of 4 in this range: 4, 8, 12, 16, 20, 24. Number of multiples of 4 = 6. Multiples of 7 in this range: 7, 14, 21. Number of multiples of 7 = 3. Common multiples of 4 and 7 would be multiples of 28. Since 28 is greater than 25, there is no common multiple in the range. Therefore, total favourable integers divisible by 4 or 7 = 6 + 3 = 9. Probability = favourable outcomes / total outcomes = 9 / 25. Verification / Alternative check: As a quick check, we can list all favourable numbers explicitly: 4, 7, 8, 12, 14, 16, 20, 21, 24. Counting them gives 9 numbers. Since the total possibilities are 25 equally likely outcomes, the fraction 9/25 is consistent and cannot be simplified further, so it is in lowest terms. Why Other Options Are Wrong: 3/25: This would mean only 3 favourable numbers, which is too small and does not match the actual list of multiples. 1/25: This would correspond to exactly one favourable number, which is clearly not true here. 7/25: This assumes only 7 favourable numbers, which does not match the actual count of 9. None of these: This is incorrect because 9/25 is a valid and correct probability among the given options. Common Pitfalls: A common mistake is to forget to check for overlaps when using the phrase “divisible by 4 or 7” and directly add the counts even when there is a common multiple. Another frequent error is miscounting multiples in the given range or accidentally including a number outside the range. Candidates may also confuse “or” with “and” and try to find numbers divisible by both 4 and 7, which is not what the question asks. Final Answer: Therefore, the required probability that the selected number is divisible by 4 or by 7 is 9/25.

Question 6

Four persons are chosen at random from a group of 3 men, 2 women, and 4 children. What is the probability that the selected group contains exactly 1 man, 1 woman, and 2 children?

Accepted Answer

Introduction / Context: This problem is a classic example of probability with combinations, where we are selecting people from different categories. The group has men, women, and children, and we want the probability that the chosen group of four contains exactly one man, one woman, and two children. This tests understanding of counting methods, combinations, and probability based on equally likely combinations. Given Data / Assumptions: Total people = 3 men + 2 women + 4 children = 9 persons. We must select exactly 4 persons. Required composition: 1 man, 1 woman, and 2 children. All combinations of 4 persons from the 9 are equally likely. Sampling is without replacement. Concept / Approach: We use combinations nCr to count the number of ways to choose people from each category. The probability is computed as (number of favourable groups) divided by (total number of possible groups of 4). We count favourable outcomes by choosing 1 man from 3, 1 woman from 2, and 2 children from 4, and we multiply these because the choices are independent categorywise. Then we divide by the total number of ways to choose 4 people from 9. Step-by-Step Solution: Total number of ways to choose 4 persons from 9 = C(9,4). Compute C(9,4) = 9 * 8 * 7 * 6 / (4 * 3 * 2 * 1) = 126. Number of ways to choose 1 man from 3 = C(3,1) = 3. Number of ways to choose 1 woman from 2 = C(2,1) = 2. Number of ways to choose 2 children from 4 = C(4,2) = 6. Total favourable groups = 3 * 2 * 6 = 36. Required probability = favourable / total = 36 / 126. Simplify 36 / 126 by dividing numerator and denominator by 18: 36 / 126 = 2 / 7. Verification / Alternative check: An alternative way is to verify that there is no other composition counted by mistake. Since the requirement is exactly 1 man, 1 woman, and 2 children, there is no overlap with any other composition such as 2 men or 3 children. Therefore, every counted group matches the requirement. The simplified fraction 2/7 is already in lowest terms, so the probability is correct. Why Other Options Are Wrong: 3/7: This would require 54 favourable groups, which is larger than the actual count of 36. 4/7: This is too large and would imply 72 favourable groups, which is impossible since the total number of groups is only 126. 5/21: This fraction is approximately 0.238, which does not match the exact ratio 36/126 = 0.2857. None of these: This is incorrect because 2/7 is a valid and correct option in the list. Common Pitfalls: A common error is to forget that we are dealing with combinations, not permutations, and to multiply in the wrong way. Another pitfall is to misinterpret the phrase "exactly 1 man, 1 woman, and 2 children" and accidentally include groups with more children or more adults. Some learners also forget to compute the total number of groups correctly and instead use an incorrect denominator, which changes the probability value completely. Final Answer: Hence, the probability of selecting exactly 1 man, 1 woman, and 2 children is 2/7.

Question 7

Two unbiased coins are tossed simultaneously. What is the probability that at most one head appears (that is, zero or one head)?

Accepted Answer

Introduction / Context: This problem focuses on basic probability with a small and simple sample space. Tossing two unbiased coins is one of the most familiar examples in probability. The question asks for the probability of getting at most one head, which means the number of heads can be zero or one, but not two. This tests understanding of sample spaces and counting favourable outcomes correctly. Given Data / Assumptions: Two unbiased (fair) coins are tossed simultaneously. Each coin has two outcomes: Head (H) or Tail (T). All four possible ordered outcomes are equally likely. We require the probability of getting at most one head (0 or 1 head). Concept / Approach: For two coin tosses, it is easiest to list all possible outcomes. Then we count how many outcomes satisfy the condition "at most one head". At most one head means the allowed numbers of heads are 0 and 1. We then divide the count of favourable outcomes by the total number of possible outcomes in the sample space. Step-by-Step Solution: List the sample space for two coin tosses: {HH, HT, TH, TT}. Total number of possible outcomes = 4. Outcomes with zero heads: TT (this has 0 heads). Outcomes with exactly one head: HT and TH (each has 1 head). So the favourable outcomes are: TT, HT, TH. Number of favourable outcomes = 3. Probability = favourable outcomes / total outcomes = 3 / 4. Verification / Alternative check: An alternative way is to recognize that the only outcome that does not satisfy the requirement is HH, which has two heads. The probability of HH is 1 out of 4. Therefore, the probability of at most one head is 1 minus the probability of two heads: 1 - 1/4 = 3/4. This matches the previous calculation exactly. Why Other Options Are Wrong: 1/2: This would imply that half the outcomes are favourable, but we have 3 favourable outcomes out of 4, not 2 out of 4. 3/2: This is greater than 1 and therefore cannot be a probability value at all. 1/6: This is much too small and does not reflect the actual count of favourable outcomes in a four outcome space. None of these: This is incorrect because 3/4 is present and is correct. Common Pitfalls: Many learners misinterpret the phrase "at most one head" and think it means exactly one head, missing the zero head case. Another mistake is to forget that the order of outcomes matters in the sample space, which always has four outcomes for two unbiased coins. Also, probabilities greater than 1 should be immediately rejected, since they are invalid in any probability problem. Final Answer: Thus, the probability that at most one head appears when two unbiased coins are tossed is 3/4.

Question 8

A coin is tossed and a fair six-sided die is rolled. What is the probability of obtaining a tail on the coin and a 4 on the die in the same trial?

Accepted Answer

Introduction / Context: This problem combines two simple random experiments: tossing a coin and rolling a fair six-sided die. We are interested in the probability of a specific combined event, namely getting a tail on the coin and a 4 on the die at the same time. This tests understanding of independent events and how to multiply probabilities. Given Data / Assumptions: The coin is fair and has two outcomes: Head (H) and Tail (T). The die is fair with six faces numbered 1 to 6. The coin toss and the die roll are independent events. We want the probability of the combined event: T on the coin and 4 on the die. Concept / Approach: When two events are independent, the probability of both occurring together is the product of their individual probabilities. Therefore, we find P(T) for the coin and P(4) for the die, then multiply them. Alternatively, we can think in terms of an overall sample space of 2 * 6 = 12 equally likely outcomes and count the favourable outcome directly. Step-by-Step Solution: The probability of getting a tail on the coin, P(T) = 1/2. The probability of getting a 4 on the die, P(4) = 1/6. Since the two experiments are independent, P(T and 4) = P(T) * P(4). Compute: P(T and 4) = (1/2) * (1/6) = 1/12. Hence, the required probability is 1/12. Verification / Alternative check: We can construct the sample space explicitly. All possible ordered pairs (coin outcome, die outcome) are: (H,1), (H,2), ..., (H,6), (T,1), (T,2), ..., (T,6). There are 12 equally likely outcomes in total. Only one of these is (T,4), which corresponds to a tail and a 4. Therefore, probability = 1 favourable outcome / 12 total outcomes = 1/12, confirming the earlier calculation. Why Other Options Are Wrong: 1/2: This might be chosen by someone who only considers the coin and ignores the die, but it does not account for the combined event. 2/3 and 3/4: These values are very large and would mean that the event happens most of the time, which is clearly not reasonable for a specific combined outcome. None of these: This is incorrect because 1/12 is one of the listed options and is correct. Common Pitfalls: A common mistake is to add probabilities instead of multiplying them for independent events. Another error is to forget that the die has six outcomes and to assume some incorrect denominator. Some learners also confuse the phrase "and" with "or" in probability and calculate the probability of tail or 4, which is a completely different event. Final Answer: Therefore, the probability of obtaining a tail on the coin and a 4 on the die in the same trial is 1/12.

Question 9

Tickets are numbered consecutively from 101 to 350 and placed in a box. If one ticket is selected at random, what is the probability that its hundreds digit is 2?

Accepted Answer

Introduction / Context: This problem deals with place value and probability in a simple discrete setting. Tickets are numbered from 101 to 350, and we draw one ticket at random. The question asks for the probability that the hundreds digit of the selected ticket is 2. This tests counting ranges and understanding of digit positions in numbers. Given Data / Assumptions: Tickets are numbered from 101 to 350 inclusive. All tickets are equally likely to be selected. We are interested in numbers whose hundreds digit is 2. Hundreds digit refers to the digit in the hundreds place of a three digit number. Concept / Approach: We need two counts: the total number of tickets and the number of tickets whose hundreds digit is 2. Once we have both, probability is favourable tickets divided by total tickets. Numbers with hundreds digit 2 lie between 200 and 299 inclusive. We also need to confirm that this block is fully inside the overall range from 101 to 350. Step-by-Step Solution: Total tickets from 101 to 350 inclusive = 350 - 101 + 1. Compute 350 - 101 = 249, then add 1 to get 250 tickets in total. Numbers with hundreds digit 2 are all integers from 200 to 299 inclusive. Count of numbers from 200 to 299 inclusive = 299 - 200 + 1 = 100. All of these 100 numbers lie inside the given range 101 to 350. Therefore, favourable tickets = 100, total tickets = 250. Probability = favourable / total = 100 / 250 = 2 / 5. Convert 2 / 5 to decimal: 2 divided by 5 = 0.40. Verification / Alternative check: We can classify by hundreds digit: from 101 to 199 (hundreds digit 1), 200 to 299 (hundreds digit 2), and 300 to 350 (hundreds digit 3). The counts in each block are 99, 100, and 51 respectively, which sum to 250. Only the middle block has hundreds digit 2, confirming that there are exactly 100 favourable tickets. The probability 100/250 reduces to 2/5, which equals 0.40, confirming our result. Why Other Options Are Wrong: 0.25: This would correspond to 62.5 favourable tickets, which is not an integer and does not match the actual count of 100. 0.5: This would imply that half of the tickets have hundreds digit 2, but only 100 out of 250 do. 0.75: This would mean three quarters of the tickets are favourable, which is far too high. None of these: This is incorrect because 0.40 is listed and is the correct probability. Common Pitfalls: Learners sometimes forget to include both endpoints when counting inclusively, which leads to errors of plus or minus one. Another common mistake is to misinterpret the hundreds digit and instead count numbers containing digit 2 anywhere. It is important to focus only on the hundreds place. Also, mixing up decimal and fractional representations can cause confusion, so reducing to 2/5 and then converting to 0.40 is a good habit. Final Answer: Thus, the probability that a randomly selected ticket has hundreds digit 2 is 0.40.

Question 10

Two fair six-sided dice are tossed together. What is the probability that the total score shown is a prime number?

Accepted Answer

Introduction / Context: This question examines the probability that the sum of two fair dice is a prime number. It is a standard problem in aptitude tests and helps reinforce knowledge of prime numbers, enumeration of outcomes, and calculation of probability from a finite sample space. Given Data / Assumptions: Two fair six-sided dice are rolled simultaneously. Each die has faces numbered 1 to 6. There are 36 equally likely ordered outcomes (6 * 6). We want the probability that the total score (sum of the two faces) is a prime number. Concept / Approach: Prime numbers possible as sums of two dice range from 2 to 12. We first list the prime numbers in this range and then count the number of ordered pairs (i, j) that add up to each prime. The probability is the ratio of the total number of favourable ordered pairs to the total of 36 outcomes. Step-by-Step Solution: Possible sums from two dice range from 2 to 12. Prime numbers in this range are: 2, 3, 5, 7, 11. Sum 2: (1,1) gives 1 outcome. Sum 3: (1,2), (2,1) gives 2 outcomes. Sum 5: (1,4), (2,3), (3,2), (4,1) gives 4 outcomes. Sum 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) gives 6 outcomes. Sum 11: (5,6), (6,5) gives 2 outcomes. Total favourable outcomes = 1 + 2 + 4 + 6 + 2 = 15. Total possible outcomes = 6 * 6 = 36. Required probability = 15 / 36 = 5 / 12 after dividing numerator and denominator by 3. Verification / Alternative check: We can verify by observing that the sum distribution of two dice is symmetric and well known. The counts for sums 2 to 12 are 1,2,3,4,5,6,5,4,3,2,1. Selecting only the prime sums 2,3,5,7,11 and adding their counts gives 1 + 2 + 4 + 6 + 2 = 15, which is consistent with the previous enumeration. Dividing by 36 again yields 5/12, confirming the result. Why Other Options Are Wrong: 1/6: This would correspond to only 6 favourable outcomes, which does not match the actual count of 15. 1/2: This implies 18 favourable outcomes, which is larger than the actual value. 3/4: This is much too large and would mean that three quarters of all outcomes give a prime sum, which is not true. None of these: This is incorrect because 5/12 is present and correct. Common Pitfalls: Many learners forget that the dice are ordered, so (1,2) and (2,1) are two different outcomes. Another mistake is misidentifying prime numbers or including 1 as a prime. It is also easy to miscount the number of combinations that give each sum if one does not list them systematically. Careful enumeration avoids these errors. Final Answer: Therefore, the probability that the total score is a prime number is 5/12.

Question 11

A fair six-sided die is rolled twice. What is the probability that the sum of the two outcomes is equal to 9?

Accepted Answer

Introduction / Context: Here we consider two rolls of a fair six-sided die and ask for the probability that the total of the two rolls equals 9. This is a standard discrete probability problem that reinforces the method of listing combinations that lead to a particular sum and then dividing by the total number of possible ordered outcomes. Given Data / Assumptions: A fair die with faces numbered 1 to 6 is rolled twice. Each ordered pair of outcomes (first roll, second roll) is equally likely. Total number of ordered outcomes is 6 * 6 = 36. We require the probability that the sum of the two rolls is 9. Concept / Approach: The key idea is to list all pairs of integers (a, b) where a and b are between 1 and 6 and a + b = 9. Each of these is one favourable outcome. Then we divide the number of favourable outcomes by 36. Because the die is fair and the rolls are independent, this method is valid and straightforward. Step-by-Step Solution: All possible sums from two rolls range from 2 to 12. We are interested only in the sum 9. List all ordered pairs (a, b) such that a + b = 9 with 1 ≤ a, b ≤ 6. These pairs are: (3,6), (4,5), (5,4), (6,3). Number of favourable outcomes = 4. Total number of possible ordered outcomes = 36. Required probability = 4 / 36. Simplify 4 / 36 by dividing by 4 to get 1 / 9. Verification / Alternative check: We can check using the known distribution of sums for two dice. For sum 9, the known count is 4 outcomes, corresponding exactly to the pairs found: (3,6), (4,5), (5,4), (6,3). Dividing 4 by 36 gives 1/9. Since this matches both enumeration and the standard distribution table, the answer is confirmed. Why Other Options Are Wrong: 1/6: This would correspond to 6 favourable outcomes, which is larger than the actual number 4. 1/2 and 3/4: These probabilities are far too large for one specific sum, as they would mean that half or three quarters of all outcomes yield a sum of 9. None of these: This is incorrect because 1/9 is listed among the options and is correct. Common Pitfalls: A common mistake is to forget that the order of the two rolls matters, sometimes counting (3,6) and (6,3) as only one outcome. Another error is to try to compute the probability without listing pairs, leading to confusion about how many combinations actually give the sum 9. Systematic listing of pairs is the safest method in such questions. Final Answer: Thus, the probability that the sum of the two dice is 9 is 1/9.

Question 12

Events K and L are defined on the same sample space. Given that P(K) = 0.8 and P(L) = 0.6, can K and L be disjoint events?

Accepted Answer

Introduction / Context: This question tests conceptual understanding of disjoint (mutually exclusive) events in probability theory. Events are called disjoint if they cannot occur at the same time, which has a direct implication for the relationship between their probabilities. We are given P(K) and P(L) and asked whether K and L can be disjoint, so we must recall the basic rule for mutually exclusive events. Given Data / Assumptions: P(K) = 0.8. P(L) = 0.6. K and L are events on the same sample space. All probabilities must lie between 0 and 1. Concept / Approach: If two events K and L are disjoint, then P(K and L) = 0, and the probability of their union satisfies P(K or L) = P(K) + P(L). Since probabilities of any event cannot exceed 1, this sum must be less than or equal to 1 for disjoint events. Therefore, to check if K and L can be disjoint, we simply add their probabilities and see whether the result is at most 1. Step-by-Step Solution: For disjoint events, the rule is P(K or L) = P(K) + P(L). Given P(K) = 0.8 and P(L) = 0.6. Compute P(K) + P(L) = 0.8 + 0.6 = 1.4. For any event in a probability space, the probability cannot exceed 1. If K and L were disjoint, we would have P(K or L) = 1.4, which is impossible. Therefore, K and L cannot be disjoint events. Verification / Alternative check: Another way to argue is to note that for any two events, P(K or L) ≤ 1. When events are disjoint, P(K or L) equals the sum P(K) + P(L). Therefore, for disjoint events, P(K) + P(L) must be less than or equal to 1. Here, the sum is 1.4, which violates this basic requirement. So we can be completely certain that K and L cannot be mutually exclusive and must have some common outcomes with positive probability. Why Other Options Are Wrong: Yes, they can be disjoint events: This ignores the fact that the sum of probabilities exceeds 1, which is impossible for disjoint events. They are always independent: The information given does not guarantee independence, and in fact, the large probabilities plus non disjoint condition say nothing about independence. There is not enough information to decide: There is enough information because the numerical values directly show that the disjoint condition cannot hold. Both events must be complements: Complementary events have probabilities that sum to exactly 1, not 1.4. Common Pitfalls: A frequent confusion is between independence and disjointness. Disjoint events cannot happen together, while independent events may occur together but do not influence each other in terms of probability. Another common mistake is forgetting that probabilities cannot exceed 1, which is crucial here. Always check the sum of probabilities when dealing with mutually exclusive events. Final Answer: Hence, K and L cannot be disjoint events. The correct statement is No, they cannot be disjoint events.

Question 13

A fair six-sided die is rolled twice. What is the probability of getting a 4 on the first roll and not getting a 6 on the second roll?

Accepted Answer

Introduction / Context: This question involves two successive rolls of a fair six-sided die and asks for the probability of a specific ordered event: the first roll shows 4 and the second roll shows any number except 6. It is a typical example of a probability problem with independent events. Given Data / Assumptions: A fair six-sided die with faces 1, 2, 3, 4, 5, 6 is rolled twice. Each roll is independent of the other. We are interested in the event: first roll = 4 and second roll ≠ 6. Each face has probability 1/6 on any single roll. Concept / Approach: For independent events A and B, the probability of both A and B occurring is P(A) * P(B). Here, event A is getting 4 on the first roll, and event B is not getting 6 on the second roll. We find each probability separately and multiply them. Alternatively, we can think in terms of the 36 ordered pairs of outcomes and count the favourable ones. Step-by-Step Solution: Event A: first roll is 4. P(A) = 1/6. Event B: second roll is not 6. There are 5 favourable faces (1, 2, 3, 4, 5). So P(B) = 5/6. Because the rolls are independent, P(A and B) = P(A) * P(B). Compute P(A and B) = (1/6) * (5/6) = 5/36. Therefore, the required probability is 5/36. Verification / Alternative check: We can also view this as an ordered pair problem. Total outcomes = 36 pairs (first, second). The first must be 4, giving 6 possibilities for the second roll, but we exclude the second roll equal to 6. So there are 5 favourable pairs: (4,1), (4,2), (4,3), (4,4), (4,5). Thus, favourable outcomes = 5, total outcomes = 36, giving probability 5/36, consistent with the multiplication approach. Why Other Options Are Wrong: 1/36: This would be the probability of a single specific ordered pair, such as (4,6), not of 5 different favourable pairs. 1/12 and 1/9: These values do not equal 5/36 when simplified and correspond to different counts of favourable outcomes. None of these: This is incorrect because 5/36 is provided as one of the answer choices and is correct. Common Pitfalls: A typical mistake is to confuse "not 6" with "greater than 4" or some other condition, which changes the number of favourable outcomes. Another error is to forget independence and try to treat the probabilities additively. Always remember that for independent events, the joint probability is the product of the individual probabilities. Final Answer: Hence, the probability of getting a 4 on the first roll and not getting a 6 on the second roll is 5/36.

Question 14

An unbiased six-sided die is rolled once. What is the probability that the number obtained is a multiple of 3?

Accepted Answer

Introduction / Context: This question is a straightforward example of probability with a single fair die. It asks for the probability that the outcome is a multiple of 3. Such problems help reinforce counting favourable outcomes and dividing by the total number of equally likely outcomes. Given Data / Assumptions: An unbiased six-sided die has faces numbered 1, 2, 3, 4, 5, 6. Each face is equally likely to appear on a single roll. We want the probability that the result is a multiple of 3. Concept / Approach: A multiple of 3 among the numbers 1 to 6 is any number that is divisible by 3 without remainder. We list such numbers in the set {1, 2, 3, 4, 5, 6}, count them, and then divide by 6, which is the total number of possible outcomes. Step-by-Step Solution: Possible outcomes on a six-sided die are: 1, 2, 3, 4, 5, 6. Multiples of 3 in this set are: 3 and 6. Number of favourable outcomes = 2 (namely 3 and 6). Total number of possible outcomes = 6. Required probability = favourable outcomes / total outcomes = 2 / 6. Simplify 2 / 6 by dividing numerator and denominator by 2 to get 1 / 3. Verification / Alternative check: We can check verbally: in one roll, there is a one third chance to land on 3 or 6, since there are two favourable faces out of six total. This informal reasoning matches the fraction 1/3, confirming the calculation. Why Other Options Are Wrong: 1/6: This would correspond to only one favourable outcome, but there are actually two multiples of 3. 5/6: This would imply that only one face is not a multiple of 3, which is clearly incorrect for the numbers 1 to 6. 1/2: This suggests three favourable outcomes, but there are only two. None of these: This is incorrect since 1/3 is a valid and correct option given in the list. Common Pitfalls: Some learners mistakenly treat 0 as a possible outcome on a standard die, which it is not. Others may miscount the multiples of 3 or forget to simplify the fraction 2/6. Always list the set of outcomes explicitly if there is any doubt, and remember to reduce the final fraction to its simplest form where possible. Final Answer: Therefore, the probability that the outcome is a multiple of 3 is 1/3.

Question 15

Two fair six-sided dice are rolled. What is the probability that the total score is either 5 or 6?

Accepted Answer

Introduction / Context: In this problem two fair dice are rolled and we are interested in the event that the total score is either 5 or 6. This combines the probabilities of two different sums using the idea of mutually exclusive events. It tests the ability to count combinations for specific sums and to add probabilities correctly. Given Data / Assumptions: Two fair six-sided dice are rolled simultaneously. Each die shows an integer from 1 to 6. There are 36 equally likely ordered outcomes. We require the probability that the sum is 5 or 6. Concept / Approach: We first count the number of ordered pairs that sum to 5 and then the number that sum to 6. Since the events "sum is 5" and "sum is 6" are mutually exclusive (they cannot happen together on the same trial), the total favourable outcomes are the sum of these counts. Finally, we divide by 36 to get the probability. Step-by-Step Solution: Number of ordered pairs where the sum is 5: Pairs: (1,4), (2,3), (3,2), (4,1). So there are 4 outcomes with sum 5. Number of ordered pairs where the sum is 6: Pairs: (1,5), (2,4), (3,3), (4,2), (5,1). So there are 5 outcomes with sum 6. Total favourable outcomes (sum 5 or 6) = 4 + 5 = 9. Total possible outcomes from two dice = 6 * 6 = 36. Required probability = 9 / 36. Simplify 9 / 36 by dividing numerator and denominator by 9 to get 1 / 4. Verification / Alternative check: Using the known distribution of sums for two dice, the counts for sums 5 and 6 are 4 and 5 respectively. Adding them gives 9. Dividing 9 by 36 leads again to 1/4. This matches the explicit enumeration and confirms that the probability is 0.25. Why Other Options Are Wrong: 2/14: This fraction simplifies to 1/7, which is smaller than the correct value and does not align with the count of favourable outcomes. 5/18: This would correspond to 10 favourable outcomes (because 5/18 of 36 equals 10), which is incorrect. 3/4: This is much too large and would imply that most outcomes produce a sum of 5 or 6. None of these: This is incorrect because 1/4 is a listed option and is correct. Common Pitfalls: Some learners miscount the number of ways to get a particular sum, especially by forgetting that (2,3) and (3,2) are distinct outcomes. Another common error is to average the probabilities of sum 5 and sum 6 rather than adding them, which gives a completely wrong result. Always count ordered pairs carefully and remember that mutually exclusive events have their probabilities added, not averaged. Final Answer: Therefore, the probability that the total score is 5 or 6 is 1/4.

Question 16

A card is drawn at random from a standard deck of 52 playing cards. What is the probability that the card is either a queen or an ace?

Accepted Answer

Introduction / Context: This problem involves drawing a single card from a standard deck and calculating the probability that it is either a queen or an ace. It is a typical example of probability with playing cards, testing knowledge of deck composition and use of basic probability formulas. Given Data / Assumptions: A standard deck has 52 cards. There are 4 suits and each suit has 13 cards. There are 4 queens and 4 aces in the deck. No card is both a queen and an ace, so the two events are mutually exclusive. One card is drawn at random, and all cards are equally likely to be selected. Concept / Approach: We consider two mutually exclusive events: drawing a queen and drawing an ace. The probability that the drawn card is a queen or an ace is the sum of the probabilities of these two events, because they cannot happen simultaneously on a single card. We count how many cards belong to each category and divide by 52. Step-by-Step Solution: Number of queens in the deck = 4 (one in each suit). Number of aces in the deck = 4 (one in each suit). Since no card is simultaneously a queen and an ace, there is no overlap. Total favourable cards (queen or ace) = 4 + 4 = 8. Total cards in the deck = 52. Required probability = favourable / total = 8 / 52. Simplify 8 / 52 by dividing numerator and denominator by 4 to get 2 / 13. Verification / Alternative check: We can think in fractional terms: probability of a queen is 4/52 = 1/13, probability of an ace is also 4/52 = 1/13. Because the events are mutually exclusive, probability of queen or ace is 1/13 + 1/13 = 2/13. This confirms the earlier calculation based on counting favourable cards directly. Why Other Options Are Wrong: 9/13 and 11/13: These are far too large for such a specific event and would suggest that most cards are queens or aces, which is not true. 4/13: This would imply 16 favourable cards, double the actual count of 8. None of these: This is incorrect because the correct probability 2/13 is explicitly present as one of the options. Common Pitfalls: Some learners mistakenly include kings in the count or misremember the total number of each rank in a standard deck. Another frequent error is to forget to simplify the fraction 8/52, though leaving it unsimplified does not change its value. The most important conceptual point is recognizing that queen and ace are mutually exclusive outcomes for a single card, so their probabilities add directly. Final Answer: Thus, the probability that the drawn card is a queen or an ace is 2/13.

Question 17

A couple has four children. Assuming that the probability of a boy and a girl are equal and each child's gender is independent, what is the probability that the family has at least one girl?

Accepted Answer

Introduction / Context: This question concerns a family with four children and asks for the probability of having at least one girl. It is a classic example of using complementary probability in problems involving repeated independent trials, with each trial having two equally likely outcomes (boy or girl). Given Data / Assumptions: The couple has four children. Each child is equally likely to be a boy or a girl. Gender outcomes for different children are independent. We want the probability of having at least one girl among the four children. Concept / Approach: Calculating the probability of "at least one girl" directly would require summing probabilities for one girl, two girls, three girls, and four girls, which is possible but lengthy. A more efficient approach is to use the complement: first compute the probability of having no girls (that is, all boys) and then subtract that from 1. This method is standard for "at least one" type questions. Step-by-Step Solution: Probability that one child is a boy = 1/2, and girl = 1/2. Event of interest: at least one girl among four children. Complement event: no girls among four children, that is, all four are boys. Probability that one child is a boy = 1/2. Probability that all four are boys = (1/2) ^ 4 = 1 / 16. Required probability = 1 minus probability of all boys. Compute 1 - 1/16 = 15/16. Convert 15/16 to decimal: 15 divided by 16 = 0.9375. Verification / Alternative check: We can expand the binomial distribution for four independent trials. The total probability across all possibilities (0, 1, 2, 3, 4 girls) must sum to 1. The case of 0 girls has probability (1/2)^4 = 1/16. Therefore, the sum of probabilities for 1 or more girls is 1 - 1/16, which again is 15/16. This matches the earlier calculation, confirming our result of 0.9375. Why Other Options Are Wrong: 0.0625: This is 1/16 and represents the probability of all boys, not at least one girl. 0.5: This would mean that half of all such families have at least one girl, which is too small because only a small fraction have no girls. 0.0257: This value does not correspond to any standard fraction in this context and is far from the correct value. None of these: This is incorrect because 0.9375 is present in the options and matches the correct probability. Common Pitfalls: A common error is to interpret "at least one girl" as "exactly one girl". Another pitfall is to try to add many separate cases directly instead of using the complement rule. Some learners also incorrectly assume that four children automatically mean two boys and two girls, which ignores the probabilistic nature of the problem. Final Answer: Hence, the probability that the couple with four children has at least one girl is 0.9375.

Question 18

From a group of 3 men, 2 women, and 4 children (9 people in total), four persons are chosen at random. What is the probability that exactly 2 of the selected persons are children?

Accepted Answer

Introduction / Context: This is a combinatorial probability problem involving selection from different categories: men, women, and children. We are asked to find the probability that among four selected people, exactly two are children. This tests counting with combinations and careful handling of category constraints. Given Data / Assumptions: There are 3 men, 2 women, and 4 children, for a total of 9 people. Four persons are selected at random without replacement. All groups of four are equally likely. We require that exactly 2 of the selected persons are children; the remaining 2 must be adults (men or women). Concept / Approach: We first compute the total number of ways to choose any 4 persons out of 9 using combinations. Then we compute the number of favourable selections that contain exactly 2 children. For the favourable selections, we choose 2 children from 4 and 2 adults from the 5 adults (3 men + 2 women). The probability is the ratio of favourable selections to total selections. Step-by-Step Solution: Total people = 3 men + 2 women + 4 children = 9. Total number of ways to choose 4 people from 9 = C(9,4). Compute C(9,4) = 9 * 8 * 7 * 6 / (4 * 3 * 2 * 1) = 126. Number of children = 4, so ways to choose exactly 2 children = C(4,2) = 6. Number of adults (men + women) = 3 + 2 = 5. Ways to choose 2 adults from 5 = C(5,2) = 10. Total favourable groups with exactly 2 children = 6 * 10 = 60. Required probability = favourable / total = 60 / 126. Simplify 60 / 126 by dividing numerator and denominator by 6 to get 10 / 21. Verification / Alternative check: We can verify that we have not missed any case because the requirement is very clear: exactly 2 children and exactly 2 adults. There is no overlap between the count of children and adults. Double counting does not occur because we select from disjoint categories. The fraction 10/21 is in simplest form, and numerical evaluation gives approximately 0.476, which is a reasonable probability for this situation. Why Other Options Are Wrong: 9/29: This fraction does not match the ratio 60/126 and simplifies to a different decimal value. 12/21: This simplifies to 4/7, which would correspond to 72 favourable selections, not 60. 14/19: This is greater than 0.7, which is too large for such a specific composition constraint. None of these: This is incorrect because 10/21 is one of the given choices and matches the correct computation. Common Pitfalls: A typical mistake is to include cases with three or four children by accident or to forget that the remaining members after choosing children must be adults. Some learners also mistakenly treat combinations as permutations and overcount by considering order. Always remember that combinations C(n,r) count unordered selections and are appropriate for group selection problems like this one. Final Answer: Therefore, the probability that exactly 2 of the selected persons are children is 10/21.

Question 19

One card is drawn at random from a well-shuffled standard deck of 52 playing cards. What is the probability that the card drawn is red?

Accepted Answer

Introduction / Context: This is a basic probability question involving drawing a single card from a standard 52 card deck. The goal is to find the probability that the card drawn is red. It helps reinforce basic knowledge of deck composition and simple probability ratios. Given Data / Assumptions: A standard deck has 52 playing cards. There are two red suits: hearts and diamonds. Each suit has 13 cards. The deck is well shuffled and each card is equally likely to be drawn. We want the probability that the card is red (heart or diamond). Concept / Approach: Since there are two red suits out of four suits in total, exactly half of the cards in the deck are red. We can either reason directly using this symmetry or count red cards and divide by 52. Both approaches should lead to the same answer. Step-by-Step Solution: Total cards in a standard deck = 52. Number of red suits = 2 (hearts and diamonds). Each suit has 13 cards, so number of red cards = 2 * 13 = 26. Probability of drawing a red card = number of red cards / total cards. Compute probability = 26 / 52. Simplify 26 / 52 by dividing numerator and denominator by 26 to get 1 / 2. Verification / Alternative check: We can also use a symmetry argument. Two of the four suits are red, and the four suits are symmetric in the deck composition. Therefore, the chance that a random card is red is exactly half. This intuitive reasoning matches the formal calculation of 26 favourable outcomes out of 52 total outcomes, giving probability 1/2. Why Other Options Are Wrong: 1/3: This would imply roughly 17 or 18 red cards out of 52, which is not correct. 1/4: This suggests only one suit or 13 cards are red, but there are actually two red suits. 1/6: This is much too small and does not reflect the deck structure. None of these: This is incorrect because 1/2 is a listed option and is correct. Common Pitfalls: Some learners confuse the number of suits with the number of colors and incorrectly think there is only one red suit. Others may misremember the total number of cards in a standard deck. It is important to recall that hearts and diamonds are red, clubs and spades are black, and there are 52 cards in total. From this, the probability is straightforward to compute. Final Answer: Thus, the probability that the drawn card is red is 1/2.

Question 20

A basket contains 9 red balls, 7 white balls, and 4 black balls. Two balls are drawn at random without replacement. What is the probability that exactly one of the two balls drawn is white?

Accepted Answer

Introduction / Context: This question deals with drawing balls from a basket without replacement and asks for the probability that exactly one of the two drawn balls is white. It combines counting methods and basic probability rules, and is a common type of question in aptitude tests involving combinations and conditional reasoning. Given Data / Assumptions: Number of red balls = 9. Number of white balls = 7. Number of black balls = 4. Total balls in the basket = 9 + 7 + 4 = 20. Two balls are drawn at random without replacement. We want the probability that exactly one of the two balls is white. Concept / Approach: There are two equivalent approaches. One is to use combinations directly: count the number of ways to choose exactly one white ball and one non white ball, then divide by the total number of ways to choose any two balls. The other is to use ordered probabilities (drawing white then non white, plus non white then white) and sum them. The combination method is simpler and avoids double counting. Step-by-Step Solution: Total number of balls = 20. Total number of ways to choose 2 balls from 20 = C(20,2) = 20 * 19 / 2 = 190. We want exactly one white ball in the pair. Number of white balls = 7, number of non white balls (red or black) = 20 - 7 = 13. Number of ways to choose 1 white = C(7,1) = 7. Number of ways to choose 1 non white = C(13,1) = 13. Total favourable pairs with exactly one white = 7 * 13 = 91. Required probability = favourable / total = 91 / 190. Verification / Alternative check: Using the ordered probability approach, the event "exactly one white" can occur in two ways: white first then non white, or non white first then white. The probability of white then non white is (7/20) * (13/19). The probability of non white then white is (13/20) * (7/19). Adding these gives (7 * 13 / (20 * 19)) + (13 * 7 / (20 * 19)) = (2 * 7 * 13) / 380 = 182 / 380, which simplifies to 91 / 190. This matches the combination method exactly. Why Other Options Are Wrong: 18/95 and 18/190: These fractions suggest 36 favourable ordered cases or 18 unordered cases, which do not match the correct count of 91 favourable unordered pairs. 1/2: This would imply that exactly one white occurs in half of all possible pairs, which is not consistent with the given composition. None of these: This is incorrect because 91/190 is present and is the correct answer. Common Pitfalls: Some learners mistakenly compute the probability of at least one white instead of exactly one white. Others forget that the draws are without replacement and use incorrect denominators. It is also easy to double count pairs when working with ordered outcomes and then converting to unordered pairs. The combination based approach neatly avoids such issues. Final Answer: Therefore, the probability that exactly one of the two balls drawn is white is 91/190.

Probability Questions