Question 1

Three unbiased (fair) coins are tossed simultaneously. What is the probability of getting at most two heads?

Accepted Answer

Introduction / Context: This problem is related to coin toss probability and uses the phrase at most two heads, which means zero, one or two heads allowed, but not three. The easiest way is to consider the complement event of getting three heads and then subtract this probability from 1. Given Data / Assumptions: Three fair coins are tossed at the same time. Each coin has equal chance of head or tail. Total outcomes = 2^3 = 8, all equally likely. Event of interest: at most two heads (0, 1 or 2 heads). Concept / Approach: At most two heads is the complement of exactly three heads. Therefore P(at most two heads) = 1 minus P(three heads). Since each of the eight outcomes is equally likely, counting the single outcome with three heads is straightforward. This approach avoids listing all 8 outcomes in detail. Step-by-Step Solution: Total outcomes = 2^3 = 8. Outcome with three heads is HHH, so there is exactly 1 favourable outcome for three heads. P(three heads) = 1 / 8. P(at most two heads) = 1 - P(three heads) = 1 - 1 / 8 = 7 / 8. Verification / Alternative check: We can directly count outcomes with 0, 1 or 2 heads. Zero heads: TTT (1 outcome). One head: HTT, THT, TTH (3 outcomes). Two heads: HHT, HTH, THH (3 outcomes). Total with at most two heads = 1 + 3 + 3 = 7 outcomes. Probability = 7 / 8, which agrees with the complement method. Why Other Options Are Wrong: 3/4 would mean 6 favourable outcomes, but there are 7. 1/2 implies only 4 favourable outcomes out of 8, which is too small. 1/4 is the probability of getting exactly one head in some cases, not at most two heads. Common Pitfalls: A common error is misinterpreting at most two heads as exactly two heads, which would give 3/8. Others may forget to include the zero head case. Always translate at most as less than or equal to the specified number and consider using the complement if it simplifies the counting. Final Answer: Therefore, the probability of getting at most two heads is 7/8.

Question 2

A bag contains 50 tickets numbered 1, 2, 3, ..., 50. Five tickets are drawn at random and then arranged in ascending order of their numbers. What is the probability that the third ticket in this ordered list has the number 30?

Accepted Answer

Introduction / Context: This question is a more advanced combinatorial probability problem. We select five distinct numbers from 1 to 50 and then arrange them in increasing order. We are asked for the probability that the third smallest selected number is exactly 30. This requires careful thinking about how many chosen numbers lie below and above 30. Given Data / Assumptions: Tickets are numbered 1 to 50. Five distinct tickets are drawn at random. They are then arranged in ascending order: X1 < X2 < X3 < X4 < X5. We want P(X3 = 30). Concept / Approach: If the third smallest number is 30, then exactly two of the chosen numbers must be less than 30 and exactly two must be greater than 30. All choices are among distinct integers from 1 to 50. We count the number of ways to choose such a 5 element set and divide by the total number of 5 element subsets of {1, 2, ..., 50}. The ordering after selection is fixed by ascending order, so we focus on combinations. Step-by-Step Solution: Total ways to choose any 5 tickets from 50 = C(50, 5). Numbers less than 30: 1 to 29, so 29 numbers. Numbers greater than 30: 31 to 50, so 20 numbers. To have X3 = 30, choose 2 numbers from 1 to 29 and 2 numbers from 31 to 50, and include 30 itself. Ways to choose 2 from 29 = C(29, 2). Ways to choose 2 from 20 = C(20, 2). Favourable sets = C(29, 2) * C(20, 2). Total sets = C(50, 5). Probability = [C(29, 2) * C(20, 2)] / C(50, 5) = 551 / 15134. Verification / Alternative check: We can compute the numerical values: C(29, 2) = 406, C(20, 2) = 190, and C(50, 5) = 2,118,760. Then favourable sets = 406 * 190 = 77,140, and probability = 77,140 / 2,118,760. Simplifying this fraction yields 551/15134, which matches the given option. Why Other Options Are Wrong: 1/2 and 1/9 are simple fractions that do not relate to the specific combinatorial counts here. 552/15379 is a different fraction and does not arise from the correct combination values, so it must be incorrect. Common Pitfalls: Common errors include ignoring the ordering condition and not enforcing exactly two numbers below and exactly two above 30. Some learners attempt to assign probabilities to individual positions without working with combinations, which leads to incorrect reasoning. Carefully translating the order condition into a statement about how many numbers lie below and above the target value is essential. Final Answer: Therefore, the probability that the third smallest ticket is 30 is 551/15134.

Question 3

Person A speaks the truth in 75% of the cases and person B speaks the truth in 80% of the cases. If they both independently describe the same incident, what is the probability that their statements will contradict each other?

Accepted Answer

Introduction / Context: This question deals with conditional behaviour of two people describing the same incident. A speaks truth with a certain probability and B speaks truth with another probability. Their statements contradict when exactly one of them tells the truth and the other tells a lie. We model this with basic probability of independent events. Given Data / Assumptions: P(A tells truth) = 75% = 0.75, so P(A lies) = 0.25. P(B tells truth) = 80% = 0.80, so P(B lies) = 0.20. A and B speak independently. They contradict each other if one speaks truth and the other lies. Concept / Approach: Let T denote telling the truth and L denote lying. Contradiction occurs when A is T and B is L, or A is L and B is T. The total probability of contradiction is the sum of the probabilities of these two mutually exclusive cases. We use multiplication for independent events and then addition because the two cases cannot occur simultaneously. Step-by-Step Solution: P(A tells truth) = 0.75, P(A lies) = 0.25. P(B tells truth) = 0.80, P(B lies) = 0.20. Case 1: A tells truth and B lies. P(case 1) = 0.75 * 0.20 = 0.15. Case 2: A lies and B tells truth. P(case 2) = 0.25 * 0.80 = 0.20. P(contradiction) = P(case 1) + P(case 2) = 0.15 + 0.20 = 0.35. 0.35 as a percentage is 35%, that is 35/100. Verification / Alternative check: We can also compute the probability that they agree (both tell truth or both lie) and then subtract from 1. P(both truth) = 0.75 * 0.80 = 0.60. P(both lies) = 0.25 * 0.20 = 0.05. So P(agree) = 0.60 + 0.05 = 0.65. Then P(contradiction) = 1 - 0.65 = 0.35, which matches the earlier result. Why Other Options Are Wrong: 30/100 and 50/100 are different from 35/100 and do not result from the correct multiplication and addition. 45/100 would require a different set of truth and lie probabilities and is not supported by the given data. Common Pitfalls: One common mistake is to add probabilities 0.75 and 0.80 directly, ignoring independence and the need to consider separate outcome combinations. Another error is to forget that both truth and both lies lead to agreement, not contradiction. Drawing a simple table of four outcomes (TT, TL, LT, LL) helps keep the logic clear. Final Answer: Therefore, the probability that A and B contradict each other is 35/100.

Question 4

One bag contains 4 red balls and 3 black balls. A second bag contains 2 red balls and 4 black balls. One of the two bags is selected at random and then one ball is drawn at random from the chosen bag. What is the probability that the ball drawn is r…

Accepted Answer

Introduction / Context: This question combines conditional probability with a simple selection between two bags. We choose one bag at random, then draw a ball from that bag and want the probability that the ball is red. The problem naturally leads to the use of the law of total probability over the two possible bag choices. Given Data / Assumptions: Bag 1: 4 red balls, 3 black balls, total 7. Bag 2: 2 red balls, 4 black balls, total 6. One bag is selected at random, so P(Bag 1) = P(Bag 2) = 1/2. From the selected bag, one ball is drawn at random. Event of interest: the drawn ball is red. Concept / Approach: Let R be the event that the ball drawn is red. Let B1 be the event of choosing Bag 1 and B2 be the event of choosing Bag 2. Using total probability, P(R) = P(B1) * P(R | B1) + P(B2) * P(R | B2). We compute the conditional probabilities by simple fractions within each bag, then average them according to bag selection probabilities. Step-by-Step Solution: P(B1) = 1/2, P(B2) = 1/2. In Bag 1, red balls = 4 out of 7, so P(R | B1) = 4 / 7. In Bag 2, red balls = 2 out of 6, so P(R | B2) = 2 / 6 = 1 / 3. Apply total probability: P(R) = (1/2) * (4/7) + (1/2) * (1/3). P(R) = 1/2 * (4/7 + 1/3) = 1/2 * (12/21 + 7/21) = 1/2 * (19/21). P(R) = 19 / 42. Verification / Alternative check: We can think in terms of a two step experiment tree: first branch for bag choice, second for colour. The total probability of red equals the sum of probabilities along the two red ending branches: (1/2 * 4/7) and (1/2 * 2/6). Summing these again yields 19/42. No other branches lead to red, so this covers all possibilities. Why Other Options Are Wrong: 23/42 and 16/39 do not reflect any natural weighting of the two bag probabilities. 7/32 uses a denominator that suggests 32 equally likely outcomes, which is not the case here. Common Pitfalls: Some learners average the red proportions as (4/7 + 2/6) / 2 correctly but then forget this corresponds exactly to the total probability formula. Others mistakenly treat the total balls as 13 and attempt a direct ratio 6/13, which ignores the initial random bag choice. Always condition on which bag is chosen in such problems. Final Answer: Hence, the probability that the ball drawn is red is 19/42.

Question 5

From a set of 26 balls labeled with the letters A to Z, one ball is selected at random. What is the probability that the chosen ball corresponds to a vowel letter (A, E, I, O or U)?

Accepted Answer

Introduction / Context: The question tests a very basic probability idea where all outcomes are equally likely. We have balls labeled with letters of the English alphabet, and we want the probability that a randomly selected ball shows a vowel. Such questions help learners get comfortable with counting favorable outcomes and dividing by the total number of equally likely outcomes. This kind of problem is often used as the starting point for more advanced probability ideas. Given Data / Assumptions: There are 26 balls in total, each labeled with a distinct letter from A to Z. The vowel letters considered are A, E, I, O and U, so there are 5 vowels. One ball is chosen at random and every ball is equally likely to be selected. No additional structure such as replacement is involved, because only one ball is drawn. Concept / Approach: The core concept is classical probability. For equally likely outcomes, the probability of an event is defined as favorable outcomes divided by total outcomes. Here, favorable outcomes correspond to balls that carry a vowel letter. Total outcomes correspond to all available balls. We only need to count how many balls correspond to vowels and how many balls there are in total. Step-by-Step Solution: Step 1: Count the total number of balls, which is 26.Step 2: Count the vowel letters among A to Z. The vowels are A, E, I, O and U, which gives 5 vowels.Step 3: The number of favorable outcomes is therefore 5, one for each vowel labeled ball.Step 4: Use the probability formula probability = favorable outcomes / total outcomes.Step 5: Substitute values to get probability = 5 / 26.Step 6: There is no need to simplify further because 5 and 26 have no common factor other than 1. Verification / Alternative check: We can verify the count by listing vowels explicitly. The alphabet has letters A, B, C, D, E, F, G, H, I, J, K, L, M, N, O, P, Q, R, S, T, U, V, W, X, Y and Z. Among these, A, E, I, O and U are exactly 5 vowels. All letters are distinct, and each ball is equally likely, so the classical probability approach is valid and leads naturally to 5/26. There is no other subtle condition in the problem, so the calculation is consistent. Why Other Options Are Wrong: 21/26 is the probability of selecting a consonant, not a vowel. 10/26 corresponds to double counting the vowels, which has no justification in the scenario. 1/26 would be the probability if there were only one vowel, which is incorrect. 11/26 does not match any reasonable count of vowels or consonants and does not arise from correct counting. Common Pitfalls: Students sometimes forget that there are 5 vowels in the English alphabet and may accidentally include Y, which would change the count. Another common error is to mix up the probability of selecting a vowel with the probability of selecting a consonant and pick the complement instead. Learners may also try to overcomplicate this very simple scenario by imagining multiple draws when only one ball is selected. Final Answer: The correct probability that a randomly selected ball shows a vowel letter is 5/26.

Question 6

A word consists of 9 letters in total, of which 5 are consonants and 4 are vowels. Three letters are chosen at random without replacement. What is the probability that more than one vowel (that is, at least two vowels) will be selected?

Accepted Answer

Introduction / Context: This question checks understanding of selections without replacement and how to calculate probabilities involving phrases like more than one vowel. It requires counting favorable combinations of letters in a situation where the composition of the set is known. The student must carefully interpret more than one vowel as meaning at least two vowels and use basic combinatorics to count these cases. Given Data / Assumptions: Total letters in the word = 9. Consonants = 5 and vowels = 4. Three letters are selected at random without replacement. Each three letter selection is equally likely. More than one vowel means exactly two vowels or exactly three vowels. Concept / Approach: The total number of ways to choose 3 letters from 9 is computed using combinations. Favourable outcomes occur in two disjoint cases: exactly two vowels and one consonant, or exactly three vowels. We calculate the number of combinations in each case, add them, and divide by the total number of combinations. This is a standard example of using combination formulas in probability. Step-by-Step Solution: Step 1: Compute total number of ways to choose 3 letters from 9 letters: C(9,3) = 84.Step 2: For exactly two vowels and one consonant, choose 2 vowels from 4 and 1 consonant from 5. That gives C(4,2) * C(5,1) = 6 * 5 = 30 ways.Step 3: For exactly three vowels, choose 3 vowels from 4. That gives C(4,3) = 4 ways.Step 4: Total favorable selections for more than one vowel is 30 + 4 = 34.Step 5: Probability = favorable outcomes / total outcomes = 34 / 84.Step 6: Simplify 34 / 84 by dividing numerator and denominator by 2 to get 17 / 42. Verification / Alternative check: To verify, we may compute the probabilities of the complementary events (zero vowels and exactly one vowel) and check that all probabilities add to 1. Zero vowels means all three letters are consonants, which has C(5,3) = 10 ways. Exactly one vowel gives C(4,1) * C(5,2) = 4 * 10 = 40 ways. Adding favorable cases for more than one vowel (34) plus 10 and 40 gives a total of 84, which matches C(9,3), confirming that our counting is consistent. Why Other Options Are Wrong: 13/42 would correspond to a smaller number of favorable outcomes than we actually count and does not match the combination counts. 5/42 is even smaller and clearly incorrect after checking all cases. 11/42 is not supported by any natural breakdown of cases. 1/2 would suggest 42 favorable outcomes out of 84, which contradicts the exact count of 34 favorable selections. Common Pitfalls: Students sometimes misinterpret more than one vowel as exactly two vowels and forget the three vowel case. Another frequent mistake is to use permutations instead of combinations, which would overcount arrangements that differ only in order. It is also easy to miscount by accidentally using 9 choose 2 or other incorrect totals. Careful listing of cases and use of combination formulas helps avoid these errors. Final Answer: The probability that more than one vowel will be selected is 17/42.

Question 7

Three houses are available in a locality and three different persons each independently apply for one house, choosing without consulting the others. What is the probability that all three applicants apply for exactly the same house?

Accepted Answer

Introduction / Context: This problem illustrates probability with independent choices among a small number of options. Each of three persons chooses a house from three available houses, and we are interested in the chance that all three people end up choosing the same house. Such questions relate directly to the idea of outcomes in a discrete sample space where all choices are equally likely and independent. Given Data / Assumptions: There are exactly 3 distinct houses available. There are 3 distinct persons applying for houses. Each person chooses one house only. Each person chooses independently and each house is equally likely to be chosen by any person. We want the probability that all three choose the same house. Concept / Approach: The key concept is to model each person's choice as an independent event with three equally likely outcomes. The total number of possible ordered triples of choices is 3^3. We then count the favorable outcomes in which all three chosen houses are identical. Finally, we divide the favorable count by the total count to obtain the required probability. Step-by-Step Solution: Step 1: Represent each person's choice as one of three houses, say H1, H2 or H3.Step 2: Since there are 3 independent choices and each has 3 outcomes, the total number of possible ordered outcomes is 3 * 3 * 3 = 3^3 = 27.Step 3: For all three applicants to choose the same house, they may all choose H1, all choose H2, or all choose H3.Step 4: Thus, the favorable ordered triples are (H1, H1, H1), (H2, H2, H2) and (H3, H3, H3).Step 5: The number of favorable outcomes is therefore 3.Step 6: Probability = favorable outcomes / total outcomes = 3 / 27 = 1 / 9. Verification / Alternative check: We can verify this answer by checking the complement event and ensuring probabilities sum correctly. The complement event is that the three applicants are not all choosing the same house. Its probability is 1 minus our answer, which is 1 minus 1/9 = 8/9. There is no conflict with any other probabilities because all 27 outcomes are accounted for and the sample space is uniform. Why Other Options Are Wrong: 2/9 would imply 6 favorable outcomes, but only three same house outcomes exist. 8/9 is the probability that they are not all in the same house, not the event in question. 7/9 also does not correspond to any natural count in this context. 1/3 would require 9 favorable outcomes, which is not supported by any reasonable enumeration of possibilities. Common Pitfalls: A common mistake is to treat the persons as indistinguishable, which leads to undercounting the total outcomes. Another error is to think that the first person fixes the house, and then to incorrectly count subsequent probabilities without considering all three houses symmetrically. It is safer to count all ordered triples and then identify favorable ones systematically. Final Answer: The required probability that all three applicants apply for the same house is 1/9.

Question 8

The probabilities that a student will receive an A, B, C or D grade in a certain course are 0.4, 0.3, 0.2 and 0.1 respectively. What is the probability that the student will receive at least a B grade (that is, either an A or a B)?

Accepted Answer

Introduction / Context: This problem involves discrete probabilities for grade outcomes in a course. The student can receive a grade of A, B, C or D, each with specified probabilities. The question asks for the probability of receiving at least a B grade, which means the student gets either an A or a B. This introduces the idea of combining probabilities of multiple events using addition when the events are disjoint. Given Data / Assumptions: Possible grades considered are A, B, C and D. Probability of A grade = 0.4. Probability of B grade = 0.3. Probability of C grade = 0.2. Probability of D grade = 0.1. These grades represent all possible outcomes for the course. Concept / Approach: The phrase at least a B grade means that the grade must be A or B. These two events are mutually exclusive, since the student cannot receive both grades at once. Therefore, we use the addition rule for disjoint events: the probability of A or B is the sum of the probabilities of A and B. No conditional probability or complex counting is required because the outcomes are already described with explicit probabilities. Step-by-Step Solution: Step 1: Identify the event of interest as receiving at least a B grade.Step 2: This event consists of two simple outcomes: getting an A or getting a B.Step 3: From the data, P(A) = 0.4 and P(B) = 0.3.Step 4: Because A and B are mutually exclusive, the probability of A or B is P(A) + P(B).Step 5: Add these to get 0.4 + 0.3 = 0.7.Step 6: Therefore, the probability that the student will receive at least a B grade is 0.7. Verification / Alternative check: We can check that the probabilities of all grade outcomes sum to 1, confirming that the distribution is valid. Adding 0.4, 0.3, 0.2 and 0.1 gives 1.0. The probability of not receiving at least a B grade is the probability of receiving either a C or a D, which is 0.2 + 0.1 = 0.3. The complement relationship P(at least B) + P(below B) = 0.7 + 0.3 = 1 confirms consistency with the full distribution. Why Other Options Are Wrong: 0.21 is much smaller than the required 0.7 and does not correspond to adding any useful subset of given probabilities. 0.3 is the probability of receiving exactly a B grade, not at least a B grade. 0.5 corresponds to A or C, which is not the event described. 0.6 would require some other combination such as A plus C, and is not related to the phrase at least B grade. Common Pitfalls: Students sometimes misinterpret at least B as exactly B and use only 0.3. Another mistake is to include C in the sum, possibly due to confusion about grade ordering. It is also important to ensure that the events being added are disjoint; if overlapping events are added without correction, the total probability can exceed 1, which is impossible. Final Answer: The probability that the student receives at least a B grade is 0.7.

Question 9

A box contains 10 bulbs in total, of which exactly 3 are defective and the remaining 7 are good. If a random sample of 5 bulbs is drawn without replacement, what is the probability that the sample contains exactly one defective bulb?

Accepted Answer

Introduction / Context: This question is a classic example of probability without replacement, which is modeled by the hypergeometric distribution. The bulb box contains both defective and good bulbs, and we draw several bulbs simultaneously. We are specifically interested in the case where the sample of five bulbs contains exactly one defective bulb, which requires counting combinations rather than simple single draw probabilities. Given Data / Assumptions: Total number of bulbs in the box = 10. Defective bulbs = 3. Good bulbs = 7. We draw 5 bulbs at random without replacement. Every set of 5 bulbs is equally likely to be chosen. Concept / Approach: The sample space consists of all possible combinations of 5 bulbs that can be drawn from 10 bulbs. This is counted using combinations. For the favorable outcomes, we require exactly one defective bulb and the remaining four bulbs to be good. That means we choose 1 defective bulb out of 3 and 4 good bulbs out of 7. The probability is the ratio of favorable combinations to total combinations, which is a straightforward application of the hypergeometric formula. Step-by-Step Solution: Step 1: Compute the total number of ways to choose 5 bulbs from 10 bulbs: C(10,5) = 252.Step 2: To have exactly one defective bulb, choose 1 defective from 3, which gives C(3,1) = 3 ways.Step 3: For the remaining 4 bulbs, choose them from the 7 good bulbs, which gives C(7,4) = 35 ways.Step 4: The total number of favorable samples with exactly one defective bulb is 3 * 35 = 105.Step 5: The required probability is favorable combinations divided by total combinations, so probability = 105 / 252.Step 6: Simplify 105 / 252 by dividing numerator and denominator by 21 to obtain 5 / 12. Verification / Alternative check: A useful check is to verify that probabilities for all possible numbers of defective bulbs in a sample of five add up to 1. One can compute probabilities for zero defective, one defective, two defective and three defective bulbs using similar combination counts and confirm that the sum is 1. Once that is verified, the specific value 5/12 stands as the correct probability for exactly one defective bulb. Why Other Options Are Wrong: 7/12 would be the complement if the other probabilities aligned in a particular way, but detailed calculation shows that it overestimates the chance of exactly one defective. 3/14 and 1/12 correspond to much smaller favorable counts, which do not match the computed value of 105 favorable samples. 11/42 does not arise from any consistent hypergeometric calculation for this setup and therefore cannot be correct. Common Pitfalls: Students often mistakenly multiply simple probabilities like 3/10 and 7/9 multiple times instead of using combinations, which does not correctly handle the requirement for exactly one defective bulb. Another frequent issue is forgetting that the order of selection does not matter in this context, leading to overcounting. It is essential to use combinations carefully and ensure that the favorable and total counts match the actual sampling process without replacement. Final Answer: The probability that the sample of five bulbs contains exactly one defective bulb is 5/12.

Question 10

Two fair six faced dice are thrown simultaneously. What is the probability that the product of the two numbers obtained on the upper faces is an even number?

Accepted Answer

Introduction / Context: This question deals with the outcome of rolling two standard dice and examines the parity of the product of the two numbers. Instead of counting all even products directly, it is usually simpler to consider when the product is odd and then take the complement. This illustrates how complementary events can make calculations easier in basic probability problems. Given Data / Assumptions: Two fair six sided dice are rolled simultaneously. Each die can show any integer from 1 to 6. All 36 ordered pairs of outcomes are equally likely. We need the probability that the product of the two numbers is even. Concept / Approach: A product of two integers is odd only when both integers are odd. If either number is even, then the product is even. Hence, the easiest approach is to first find the probability that both numbers are odd and therefore the product is odd, and then subtract that probability from 1 to obtain the probability that the product is even. This uses the complement rule in probability. Step-by-Step Solution: Step 1: Each die has 3 odd faces, which are 1, 3 and 5, and 3 even faces, which are 2, 4 and 6.Step 2: The total number of outcomes when two dice are rolled is 6 * 6 = 36.Step 3: Count outcomes where both dice show odd numbers. There are 3 odd choices for the first die and 3 odd choices for the second die, giving 3 * 3 = 9 outcomes.Step 4: In these 9 outcomes, the product is odd. Therefore, the probability that the product is odd is 9 / 36.Step 5: Probability that the product is even is the complement: 1 minus 9 / 36 = 27 / 36.Step 6: Simplify 27 / 36 by dividing numerator and denominator by 9 to get 3 / 4. Verification / Alternative check: As an alternative, we can count even products directly. If either die is even, the product is even. We can count all combinations where at least one die is even and show that there are 27 such outcomes. Another way is to check a few specific examples like (2,1), (4,3) and (6,5) to confirm that the presence of at least one even number guarantees an even product. Both approaches reach the same probability value, which validates the result. Why Other Options Are Wrong: 3/8 and 5/16 are too small and would imply that the majority of products are odd, which is not reasonable given that half of the faces on each die are even. 2/7 does not simplify nicely in relation to powers of 6 and does not match any clear counting logic. 1/4 is the probability that both numbers are even only under certain miscalculations, not the probability that the product is even in general. Common Pitfalls: A common mistake is to suppose that even and odd products occur equally often and to guess 1/2, but the calculations show that even products are more frequent. Another error is miscounting the number of odd faces on each die or forgetting that all 36 ordered pairs are equally likely. Remembering that the complement of an event often simplifies counting is a key lesson from this problem. Final Answer: The probability that the product of the two numbers is even is 3/4.

Question 11

Two fair six faced dice are thrown once. What is the probability that the sum of the numbers appearing on the upper faces is exactly 9?

Accepted Answer

Introduction / Context: This problem is about the distribution of sums when two standard dice are rolled. Each die has faces numbered from 1 to 6, and we want the probability that the sum of the two numbers equals 9. Such questions help learners become familiar with listing pairs of outcomes and understanding how some sums are more likely than others. Given Data / Assumptions: Two fair six sided dice are rolled simultaneously. Each die can show an integer from 1 to 6. All 36 ordered pairs of outcomes are equally likely. We seek the probability that the sum of the two results equals 9. Concept / Approach: The key idea is to count the number of outcome pairs that give the desired sum. Each outcome is an ordered pair (first die, second die). We list all pairs whose components add to 9 and then divide by the total number of possible outcomes, which is 36. This direct counting method is efficient for sums in dice problems. Step-by-Step Solution: Step 1: Total possible ordered outcomes when two dice are rolled is 6 * 6 = 36.Step 2: List all pairs of numbers (x, y) with x from 1 to 6 and y from 1 to 6 such that x + y = 9.Step 3: The valid pairs are (3, 6), (4, 5), (5, 4) and (6, 3).Step 4: Count these pairs and observe that there are 4 favorable outcomes.Step 5: Probability of obtaining a sum of 9 is therefore 4 / 36.Step 6: Simplify 4 / 36 by dividing numerator and denominator by 4 to get 1 / 9. Verification / Alternative check: We can verify by noting that the most likely sums are around 7, and that sums like 2 and 12 have only one combination each. For sum 9, we reasonably expect a moderate number of combinations, and our list of 4 pairs is consistent with this pattern. Another check is to ensure that we have not missed any pair such as (2, 7) or (7, 2), but these are impossible because each die only goes up to 6. Thus the enumeration appears complete and correct. Why Other Options Are Wrong: 1/2 and 3/4 are much too large and do not reflect the limited number of favorable outcomes compared to 36 total possibilities. 2/9 would correspond to 8 favorable outcomes, which is double our correctly counted number. 5/36 would require exactly 5 favorable pairs, but we have found only 4 valid pairs that add up to 9. Common Pitfalls: Students sometimes miscount by forgetting the ordered nature of outcomes, or by including impossible combinations such as 7 on a single die. Some may also forget that (4, 5) and (5, 4) are distinct outcomes when the dice are distinguishable. Being systematic in listing all possible pairs and checking the sum helps avoid such mistakes. Final Answer: The probability that the sum of the two dice equals 9 is 1/9.

Question 12

In a single throw of two fair six faced dice, what is the probability that neither a doublet (both dice showing the same number) nor a total of 8 appears on the upper faces?

Accepted Answer

Introduction / Context: This problem combines two conditions in a dice experiment. We must avoid both a doublet and a sum of 8 when two dice are thrown. Such questions reinforce understanding of how to handle combined events and how to apply the idea of excluding unwanted outcomes from the full sample space of 36 possibilities. Given Data / Assumptions: Two fair six sided dice are thrown together. Each die can show a number from 1 to 6. All 36 ordered pairs of outcomes are equally likely. A doublet is any outcome where both dice show the same number, such as (2, 2). The sum 8 consists of all ordered pairs whose total equals 8. Concept / Approach: We find the probability of the complement event by subtracting the probability of undesirable outcomes from 1. The undesirable outcomes are outcomes that are doublets, or that give a total of 8, or both. We use the union of events formula, which subtracts the overlap so that shared outcomes are not counted twice. Then we subtract this union probability from 1 to get the desired probability. Step-by-Step Solution: Step 1: Total possible outcomes from rolling two dice are 6 * 6 = 36.Step 2: Count the number of doublets. These are (1,1), (2,2), (3,3), (4,4), (5,5) and (6,6), giving 6 outcomes.Step 3: Count the outcomes that give a sum of 8. The valid ordered pairs are (2,6), (3,5), (4,4), (5,3) and (6,2), giving 5 outcomes.Step 4: Identify the overlap, that is outcomes that are both doublets and sum to 8. The only such outcome is (4,4).Step 5: Use the union count formula: total undesirable outcomes = 6 + 5 minus 1 = 10.Step 6: Probability of an undesirable outcome is 10 / 36, so the desired probability is 1 minus 10 / 36 = 26 / 36, which simplifies to 13 / 18. Verification / Alternative check: We can double check by listing the 10 undesirable outcomes explicitly and confirming that they are distinct and correctly counted. The doublets are 6 in number, and among the five sum 8 outcomes, only one overlaps. Removing that single common case from the total of 11 distinct descriptions gives 10 unique undesirable outcomes. Dividing 10 by 36 gives a probability of 5/18 for the union, so its complement is indeed 13/18. Both ways lead to the same result, confirming correctness. Why Other Options Are Wrong: 7/15 and 5/18 come from incorrect counting or misinterpretation of the union and intersection. 3/16 is far from any fraction with denominator 36 and does not correspond to a natural count. 11/18 would result from subtracting 7/36 instead of 10/36, which would clearly overlook some outcomes. Common Pitfalls: A frequent mistake is to add the counts of doublets and sum 8 outcomes without subtracting the overlap, which double counts the outcome (4,4). Another error is to misidentify the pairs that sum to 8, perhaps including impossible pairs such as (1,7). Remember that each die only has faces from 1 to 6, and that union counts require careful attention to shared outcomes. Final Answer: The probability that neither a doublet nor a total of 8 appears is 13/18.

Question 13

In a box there are 8 red balls, 7 blue balls and 6 green balls, for a total of 21 balls. One ball is picked at random. What is the probability that the chosen ball is neither red nor green, that is, that it is blue?

Accepted Answer

Introduction / Context: This question deals with a simple color based probability scenario. The box contains balls of three different colors, and we are interested in the probability that a randomly chosen ball is neither red nor green. Since the only remaining color is blue, the problem effectively asks for the probability of drawing a blue ball from the mixture. Given Data / Assumptions: Number of red balls = 8. Number of blue balls = 7. Number of green balls = 6. Total number of balls = 8 + 7 + 6 = 21. One ball is drawn at random and all balls are equally likely to be selected. Concept / Approach: The event of interest is that the ball is neither red nor green. Since the only other color present is blue, this is simply the event that the ball is blue. The probability is found by dividing the number of blue balls by the total number of balls in the box. There is no conditioning or replacement involved because the selection consists of a single draw. Step-by-Step Solution: Step 1: Compute the total number of balls: 8 red + 7 blue + 6 green = 21 balls.Step 2: Identify the number of favorable outcomes, which is the number of blue balls. There are 7 blue balls.Step 3: The event that a ball is neither red nor green is exactly the same as the event that it is blue.Step 4: Use the formula probability = favorable outcomes / total outcomes.Step 5: Substitute values to get probability = 7 / 21.Step 6: Simplify 7 / 21 by dividing both numerator and denominator by 7 to obtain 1 / 3. Verification / Alternative check: We can verify this answer by also computing the probabilities of drawing red or green balls and ensuring that all probabilities sum to 1. The probability of red is 8 / 21, of blue is 7 / 21 and of green is 6 / 21. Adding them gives (8 + 7 + 6) / 21 = 21 / 21 = 1. This confirms that our probability distribution is consistent and that 1 / 3 for blue is correct. Why Other Options Are Wrong: 3/5 would require 12 or 13 favorable outcomes out of 21, which is inconsistent with the count of 7 blue balls. 8/21 corresponds to the probability of drawing a red ball, not a blue one. 7/21 is the unsimplified form of the correct answer, but 1/3 is the reduced and standard form. 2/7 does not correspond to any natural color count in this scenario, and therefore cannot be right. Common Pitfalls: Students sometimes misread neither red nor green and incorrectly attempt to exclude blue instead. Another mistake is to forget to simplify the fraction, although leaving it unsimplified still represents the same probability. It is helpful to systematically list the counts of each color and then directly relate the desired event to one of those counts. Final Answer: The probability that the chosen ball is neither red nor green, that is, that it is blue, is 1/3.

Question 14

Four fair six faced dice are thrown simultaneously. What is the probability that all four dice show exactly the same number on their upper faces?

Accepted Answer

Introduction / Context: This question explores probability in a situation involving four dice instead of the more common one or two dice. We want the probability that all four dice show the same face. This reinforces the idea of independent events and counting total outcomes when multiple identical random devices are used. Given Data / Assumptions: Four fair six sided dice are thrown at the same time. Each die can land showing any integer from 1 to 6. All 6^4 = 1296 ordered combinations of faces are equally likely. We are interested in outcomes where all four dice show exactly the same number. Concept / Approach: Each die acts independently, so the total number of possible outcomes is the product of the possibilities for each die. For the favorable outcomes, all dice must show the same face. Since that face could be 1, 2, 3, 4, 5 or 6, we count the outcomes where all four dice show 1, all show 2, and so on. The probability is the ratio of these favorable outcomes to the total number of outcomes. Step-by-Step Solution: Step 1: Each die has 6 possible faces, so for four dice the total number of ordered outcomes is 6^4 = 6 * 6 * 6 * 6 = 1296.Step 2: For an outcome where all dice are the same, the first die can show any of the 6 faces.Step 3: Once the face value is chosen, all four dice must show that same value, so there is exactly 1 such ordered outcome for each chosen face.Step 4: Therefore, the number of favorable outcomes is 6, one for each possible common face value.Step 5: Probability that all four dice show the same number is 6 / 1296.Step 6: Simplify 6 / 1296 by dividing both numerator and denominator by 6 to get 1 / 216. Verification / Alternative check: We can also reason from the perspective of conditional probabilities. Suppose the first die shows some value. The probability that the second die matches it is 1/6, and the same is true for the third and fourth dice. The probability that all three remaining dice match the first is (1/6) * (1/6) * (1/6) = 1/216. This matches the result obtained from counting favorable and total outcomes, confirming the correctness of our answer. Why Other Options Are Wrong: 1/36 would be correct for two dice showing the same face but is far too large for four dice. 2/216 and 4/216 are partial counts that would correspond to only some of the possible common face values and fail to represent the full set of 6 favorable outcomes. 1/1296 would be the probability of one specific ordered outcome, such as all dice showing 1, not the event where they all show the same number regardless of which number that is. Common Pitfalls: Students sometimes forget to count all six possible identical outcomes and instead count only one specific case. Another issue is confusing ordered outcomes with unordered ones, but in this problem, the order of dice does matter when counting the total outcomes. Understanding that independence leads to multiplying the number of possibilities for each die is crucial for this type of question. Final Answer: The probability that all four dice show exactly the same number is 1/216.

Question 15

Let Z be a standard normal random variable with mean 0 and variance 1. What is the probability that Z lies between 0 and 3.01, that is, 0 less than Z less than 3.01?

Accepted Answer

Introduction / Context: This question involves the standard normal distribution, which is a continuous probability distribution with mean 0 and variance 1. The variable Z denotes a standard normal random variable. We are asked to find the probability that Z takes a value between 0 and 3.01. Such problems require the use of standard normal probability tables or a calculator that can evaluate cumulative distribution values. Given Data / Assumptions: Z is a standard normal random variable. Mean of Z is 0 and standard deviation is 1. We are interested in P(0 less than Z less than 3.01). Standard normal tables typically give cumulative probabilities from negative infinity up to a given Z value. The standard normal distribution is symmetric around zero. Concept / Approach: To find P(0 less than Z less than 3.01), we can use the cumulative distribution function values. The expression equals F(3.01) minus F(0), where F denotes the standard normal cumulative distribution function. Because of symmetry, F(0) equals 0.5. The value F(3.01) is very close to 1 and is found from a standard table or calculator. Subtracting 0.5 from that cumulative value gives the required probability between 0 and 3.01. Step-by-Step Solution: Step 1: Express the desired probability as P(0 less than Z less than 3.01) = F(3.01) minus F(0).Step 2: Recall that for the standard normal distribution, F(0) = 0.5 because the distribution is symmetric around zero.Step 3: Use a standard normal table or calculator to find F(3.01). The value is approximately 0.9987.Step 4: Substitute these values into the expression to get P(0 less than Z less than 3.01) = 0.9987 minus 0.5.Step 5: Perform the subtraction: 0.9987 minus 0.5 = 0.4987.Step 6: Therefore, the probability that Z lies between 0 and 3.01 is approximately 0.4987. Verification / Alternative check: As a rough check, we know that almost all of the standard normal distribution lies within three standard deviations of the mean. The probability from minus 3 to 3 is about 0.9973, so the probability from 0 to 3 should be roughly half of that, about 0.4987. Extending 3 slightly to 3.01 hardly changes the value. This back of the envelope reasoning supports the more precise calculation that uses F(3.01) and gives confidence in the result. Why Other Options Are Wrong: 0.5 would be the probability from 0 to infinity, not from 0 to 3.01. 0.9987 is the cumulative probability from negative infinity up to 3.01, not between 0 and that point. 0.1217 and 0.0013 are far too small and do not match any realistic central probability range for values between 0 and a large positive Z value like 3.01. Common Pitfalls: Students sometimes forget to subtract F(0) and instead read the cumulative probability directly from the table, which overestimates the required value. Another common error is to confuse the probability from 0 to 3 with the probability from minus 3 to 3 and not take advantage of symmetry. Carefully translating the interval into differences of cumulative probabilities is essential. Final Answer: The probability that the standard normal variable Z lies between 0 and 3.01 is approximately 0.4987.

Question 16

A number is chosen at random from the first 100 positive integers, that is, from 1 to 100 inclusive. What is the probability that the selected number is divisible by 4?

Accepted Answer

Introduction / Context: This question illustrates probability over a finite uniform sample space of integers. We choose one number at random from the first 100 positive integers and want the probability that it is divisible by 4. This involves simple counting of multiples of 4 within a specified range and then forming a probability ratio. Given Data / Assumptions: The sample space consists of integers from 1 to 100 inclusive. All 100 integers are equally likely to be selected. We are interested in numbers that are divisible by 4, that is, numbers of the form 4k with integer k. The probability is favorable outcomes divided by total outcomes. Concept / Approach: The key idea is to count how many integers between 1 and 100 are multiples of 4. These form an arithmetic progression starting from 4 up to the largest multiple less than or equal to 100. Once we know this count, we divide by 100 to obtain the probability. This is a standard method for problems involving divisibility in a finite range. Step-by-Step Solution: Step 1: The total number of possible outcomes is 100, since we are selecting from 1 to 100 inclusive.Step 2: Identify the smallest positive integer divisible by 4 in this range, which is 4.Step 3: Identify the largest integer in the range 1 to 100 that is divisible by 4. The largest such number is 100, because 4 * 25 = 100.Step 4: The multiples of 4 in this range are 4, 8, 12, and so on up to 100. They form an arithmetic progression with first term 4 and common difference 4.Step 5: The number of terms in this progression is 25, since 4 * 25 = 100.Step 6: Therefore, there are 25 favorable outcomes, and the probability that the chosen number is divisible by 4 is 25 / 100 = 1 / 4. Verification / Alternative check: We can verify by observing that every group of four consecutive integers, such as 1 to 4 or 5 to 8, contains exactly one multiple of 4. The interval from 1 to 100 consists of 25 such groups. Hence there are exactly 25 numbers divisible by 4, which confirms the earlier count. This also aligns with the simple formula range size divided by step size to count equally spaced multiples. Why Other Options Are Wrong: 2 and 1 are impossible as probabilities because they exceed the maximum possible value of 1. 1/2 would require 50 favorable outcomes, meaning every second number is a multiple of 4, which is not true. 3/10 would correspond to 30 favorable outcomes, but we have seen that there are exactly 25 multiples of 4 in the range, not 30. Common Pitfalls: Students sometimes miscount the multiples by including 0 or starting from 1 instead of 4. Others may stop at 96 and forget 100, or incorrectly count the terms in the progression. It is always helpful to check the first and last multiples of 4 within the range and then compute the count using the arithmetic progression formula to avoid off by one errors. Final Answer: The probability that the selected number is divisible by 4 is 1/4.

Question 17

Two fair six faced dice are rolled simultaneously. What is the probability that the sum of the numbers showing on the two upper faces is divisible by 3 or divisible by 4?

Accepted Answer

Introduction / Context: This question combines divisibility with dice sums. When two dice are rolled, we can obtain sums from 2 to 12. We are asked for the probability that the sum is divisible by 3 or divisible by 4. This illustrates how to handle union probabilities in a finite sample space where events may overlap, because some sums are divisible by both 3 and 4. Given Data / Assumptions: Two fair six sided dice are rolled simultaneously. Each die shows an integer from 1 to 6. There are 6 * 6 = 36 equally likely ordered outcomes. We are interested in sums that are multiples of 3 or multiples of 4. We must be careful not to double count sums that satisfy both conditions. Concept / Approach: We first list all possible sums from 2 to 12 and identify which sums are divisible by 3 or 4. We then count the number of ordered pairs of dice outcomes that yield these sums. Because each sum has a known number of generating pairs, this counting can be done by listing or by using known frequency patterns. The probability is the number of favorable outcomes divided by 36. Step-by-Step Solution: Step 1: The possible sums of two dice range from 2 to 12.Step 2: Identify sums divisible by 3: these sums are 3, 6, 9 and 12.Step 3: Identify sums divisible by 4: these sums are 4, 8 and 12.Step 4: The union of these sums is {3, 4, 6, 8, 9, 12}. Note that 12 appears in both lists.Step 5: Count the number of ordered outcomes giving each of these sums. The sum 3 has 2 outcomes, sum 4 has 3, sum 6 has 5, sum 8 has 5, sum 9 has 4 and sum 12 has 1, giving a total of 2 + 3 + 5 + 5 + 4 + 1 = 20 outcomes.Step 6: The probability that the sum is divisible by 3 or 4 is 20 / 36, which simplifies to 5 / 9 after dividing numerator and denominator by 4. Verification / Alternative check: We can verify our counts using the standard triangular distribution of two dice sums. The frequencies of sums from 2 to 12 are 1, 2, 3, 4, 5, 6, 5, 4, 3, 2 and 1. Applying these counts to the sums 3, 4, 6, 8, 9 and 12 confirms the numbers used in the calculation. Another way is to write a small table of all 36 pairs and mark which meet the divisibility conditions, but that is more tedious. The consistency of the count with known sum patterns supports our result. Why Other Options Are Wrong: 3/7 and 7/11 have denominators that do not equal 36 and do not naturally arise from any count of outcomes in this dice experiment. 6/13 is an odd fraction for this context and would correspond to 24 out of 52 if scaled, which is unrelated. 4/9 would be 16 favorable outcomes, but we have counted 20 outcomes that satisfy the given condition, making 5/9 the correct choice. Common Pitfalls: Students sometimes miscount the number of outcomes for each sum or forget to include sums that are divisible by both 3 and 4, such as 12. Another issue is accidentally counting each sum only once instead of counting all ordered pairs that yield that sum. It is important to distinguish between sums and the number of ways to achieve them when converting to probabilities. Final Answer: The probability that the sum is divisible by 3 or by 4 is 5/9.

Question 18

In a simultaneous throw of a pair of fair six faced dice, what is the probability that the total (sum) of the numbers showing on the two upper faces is strictly greater than 7?

Accepted Answer

Introduction / Context: This problem asks for the probability that the sum of two dice exceeds 7. Sums obtained when rolling two fair dice range from 2 to 12. The sums greater than 7 are 8, 9, 10, 11 and 12. This classic probability question helps students become comfortable with counting outcomes from the two dice sum distribution. Given Data / Assumptions: Two fair six sided dice are rolled at the same time. Each die can show an integer from 1 to 6. There are 6 * 6 = 36 equally likely ordered outcomes. We need the probability that the sum of the outcomes is strictly greater than 7. Concept / Approach: We count how many ordered pairs (first die, second die) produce sums of 8, 9, 10, 11 or 12. Using the known distribution of sums, we can quickly determine these counts. The probability is then calculated as the number of favorable outcomes divided by 36. This uses basic counting and the idea of equally likely outcomes in a finite sample space. Step-by-Step Solution: Step 1: The total number of ordered outcomes when rolling two dice is 36.Step 2: Identify sums greater than 7, which are 8, 9, 10, 11 and 12.Step 3: Use the standard sum distribution: sum 8 has 5 outcomes, sum 9 has 4 outcomes, sum 10 has 3 outcomes, sum 11 has 2 outcomes and sum 12 has 1 outcome.Step 4: Add these counts to find the total number of favorable outcomes: 5 + 4 + 3 + 2 + 1 = 15.Step 5: The required probability is favorable outcomes divided by total outcomes, so probability = 15 / 36.Step 6: Simplify 15 / 36 by dividing numerator and denominator by 3 to get 5 / 12. Verification / Alternative check: As an alternative, we can use the complement event. The complement of the sum being greater than 7 is the sum being less than or equal to 7. The sums 2, 3, 4, 5, 6 and 7 have 1, 2, 3, 4, 5 and 6 outcomes respectively, totaling 21 outcomes. Thus, the probability of sum less than or equal to 7 is 21 / 36, so the probability of sum greater than 7 is 1 minus 21 / 36 = 15 / 36, which again simplifies to 5 / 12. This matches our earlier result. Why Other Options Are Wrong: 1/2 would correspond to 18 favorable outcomes, which is more than the actual count of 15. 7/15 has denominator 15 and does not fit the 36 outcome space. 3/12 simplifies to 1/4, implying only 9 favorable outcomes, which is too few. 11/36 would correspond to 11 favorable outcomes, again not matching the counted 15 outcomes for sums greater than 7. Common Pitfalls: Students often misremember the counts of outcomes for each sum or assume that all sums from 2 to 12 are equally likely, which is not true. Another common error is to mistakenly include sum 7 when the condition requires strictly greater than 7. Using a table or diagram of the two dice grid helps maintain accuracy when counting. Final Answer: The probability that the sum of the two dice is greater than 7 is 5/12.

Question 19

A number X is chosen at random from the set of integers {−3, −2, −1, 0, 1, 2, 3}. What is the probability that the absolute value of X is less than 2, that is, that |X| is less than 2?

Accepted Answer

Introduction / Context: This question focuses on a finite set of integers and uses the concept of absolute value. We choose an integer X from a small symmetric set and ask for the probability that its absolute value is less than 2. This tests understanding of absolute value inequalities and basic probability over a uniform discrete sample space. Given Data / Assumptions: The set of possible values for X is {−3, −2, −1, 0, 1, 2, 3}. Each of the 7 integers is equally likely to be chosen. We need the probability that |X| less than 2. The absolute value of a number is its distance from zero on the number line. Concept / Approach: The inequality |X| less than 2 means that X lies strictly between −2 and 2. Thus the allowed integers are exactly those that have distance from zero less than 2 units. We first identify which integers in the given set satisfy this condition and then divide the number of such integers by the total number of integers in the set to get the probability. Step-by-Step Solution: Step 1: List the set of possible values of X explicitly: −3, −2, −1, 0, 1, 2 and 3.Step 2: Interpret |X| less than 2 as requiring that the distance from zero is less than 2, so X must lie strictly between −2 and 2.Step 3: The integers in the set that satisfy −2 less than X less than 2 are −1, 0 and 1.Step 4: Therefore, there are 3 favorable values of X: −1, 0 and 1.Step 5: The total number of possible values is 7, since the set has 7 elements.Step 6: The required probability is favorable outcomes divided by total outcomes = 3 / 7. Verification / Alternative check: We can check the complement event, where |X| is greater than or equal to 2. The integers with absolute value at least 2 are −3, −2, 2 and 3, which are 4 in number. Thus the probability of the complement event is 4 / 7. Adding this to the probability 3 / 7 for the original event gives 1, which confirms that our calculations account for all possible values of X correctly. Why Other Options Are Wrong: 3/4 would require that 6 out of the 7 integers satisfy the inequality, which is not true. 4/5 would indicate 5.6 favorable outcomes, which makes no sense in this discrete context. 5/7 would correspond to 5 favorable integers, but only 3 actually satisfy |X| less than 2. 2/7 would only count two values, but we have identified three integers with the required property. Common Pitfalls: Students sometimes misread |X| less than 2 as |X| less than or equal to 2 and mistakenly include −2 and 2. Others misinterpret absolute value as a signed quantity rather than distance, leading to confusion. It is also easy to overlook 0 as a valid value when listing integers, but 0 clearly has absolute value 0, which is less than 2. Final Answer: The probability that the absolute value of X is less than 2 is 3/7.

Question 20

A bag contains 2 red balls, 3 green balls and 2 blue balls. Two balls are drawn at random from the bag without replacement. What is the probability that none of the balls drawn is blue, that is, both balls are either red or green?

Accepted Answer

Introduction / Context: This question is a probability problem involving drawing balls from a bag without replacement. The bag contains balls of three different colors, and we are required to find the probability that neither of the two drawn balls is blue. This tests understanding of combinations and probabilities in the context of sampling without replacement, where order does not matter for the subset chosen. Given Data / Assumptions: Red balls in the bag = 2. Green balls in the bag = 3. Blue balls in the bag = 2. Total balls in the bag = 2 + 3 + 2 = 7. Two balls are drawn at random without replacement. We want the probability that both balls drawn are not blue. Concept / Approach: The sample space consists of all possible unordered pairs of balls drawn from the 7 balls. We use combinations to count these pairs. For the event that no drawn ball is blue, both balls must come from the 5 balls that are red or green. Hence we count combinations chosen from these 5 non blue balls. The probability is the ratio of the number of favorable pairs to the total number of pairs from all 7 balls. Step-by-Step Solution: Step 1: Compute total number of unordered ways to draw 2 balls from 7 balls: C(7,2) = 21.Step 2: Identify the number of balls that are not blue. There are 2 red and 3 green balls, giving 5 non blue balls in total.Step 3: Compute the number of unordered ways to draw 2 balls from these 5 non blue balls: C(5,2) = 10.Step 4: These 10 combinations represent all favorable outcomes where neither drawn ball is blue.Step 5: The required probability is favorable combinations divided by total combinations, so probability = 10 / 21.Step 6: This fraction is already in simplest form, so the probability that none of the two balls is blue is 10 / 21. Verification / Alternative check: We can verify using a complementary approach by finding the probability that at least one ball is blue and then subtracting from 1. The probability that the first ball is blue is 2 / 7, and conditional probabilities for the second draw depend on the first. However, the combination method is more straightforward and avoids order based complications. Checking that C(5,2) plus combinations involving blue balls equals C(7,2) confirms that we have accounted for all possibilities. Why Other Options Are Wrong: 11/21 would require 11 favorable non blue combinations, which exceeds the 10 that actually exist. 1/2 is a rough guess without considering the exact counts, and 2/7 would correspond to a much smaller set of favorable outcomes. 5/21 is half of the correct numerator and does not match any meaningful selection pattern in this context. Common Pitfalls: Students sometimes treat the draws as ordered and then forget to divide out permutations, leading to double counting. Another error is to use simple products of probabilities like (5/7) * (4/6) for non blue draws, which actually gives the same result but may be used incorrectly if the second fraction is miscalculated. Being methodical with combinations ensures that the correct result is obtained even when the numbers are more complex. Final Answer: The probability that neither of the two balls drawn is blue is 10/21.

Probability Questions