Statistics Quiz
Practice Finding Conditional Probability In Models in Statistics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.
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A class has 10 students. Each student is equally likely to be selected at random.
The students are:
Let event A be “the selected student plays soccer” and event B be “the selected student plays a sport (soccer or basketball).”
Using the list above, what is P(A∣B)? Give your answer as a simplified fraction.
This quiz focuses on Finding Conditional Probability In Models, giving you a quick way to practice the rules, question types, and explanations that matter most for Statistics.
Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.
A class has 10 students. Each student is equally likely to be selected at random.
The students are:
Let event A be “the selected student plays soccer” and event B be “the selected student plays a sport (soccer or basketball).”
Using the list above, what is P(A∣B)? Give your answer as a simplified fraction.
Explanation: This question tests conditional probability from a model of equally likely student selections. 'Given B' means we restrict our attention to only the outcomes in event B, which is 'the selected student plays a sport.' There are 7 outcomes in B: the 4 soccer players and 3 basketball players. Of these 7, 4 are also in A, which are the soccer players. Therefore, the correct fraction 4/7 represents P(A|B), as it shows the proportion of sport-playing students who play soccer. A common mistake is using the total 10 students as the denominator, leading to 4/10 instead of focusing on B. To avoid this, first list or circle the outcomes in B, then count how many of those are in A.
A fair six-sided die is rolled once. The equally likely outcomes are {1, 2, 3, 4, 5, 6}.
Let event A be “the roll is greater than 4” and event B be “the roll is even.”
Using the outcomes listed, which value best represents P(A∣B)? Give your answer as a simplified fraction.
Explanation: This question tests conditional probability from a model of equally likely die rolls. 'Given B' means we restrict our attention to only the outcomes in event B, which is 'the roll is even.' There are 3 outcomes in B: 2, 4, and 6. Of these 3, 1 is also in A, which is 6 for greater than 4. Therefore, the correct fraction 1/3 represents P(A|B), as it shows the proportion of even rolls that are greater than 4. A common mistake is using the total 6 outcomes as the denominator, leading to 2/6 instead of focusing on B. To avoid this, first circle the outcomes in B, then count how many of those are in A.
A card is chosen at random from the 16 equally likely cards labeled: A1, A2, A3, A4, B1, B2, B3, B4, C1, C2, C3, C4, D1, D2, D3, D4.
Event B: the letter is C or D. Event A: the number is 1.
Based on the outcomes listed, which value best represents P(A∣B)? Give your answer as a simplified fraction.
Explanation: This question tests conditional probability using a card model with 16 equally likely labels. Given event B, we restrict attention to letters C or D: C1-C4 and D1-D4, totaling 8 outcomes. Of these, the ones in A with number 1 are C1 and D1, so 2 outcomes. The fraction 2/8 simplifies to 1/4, representing P(A|B) as the proportion of C or D cards with number 1. A common mistake is using the total 16 as the denominator, giving 4/16 which simplifies similarly but misses the restriction. To avoid errors, circle the B group first, then count A within it for the proper conditional probability.
A fair coin is flipped twice. The 4 equally likely outcomes are: HH, HT, TH, TT.
Event B: at least one flip is H. Event A: the two flips match (HH or TT).
From the outcomes listed, what is P(A∣B)? Give your answer as a simplified fraction.
Explanation: This question tests conditional probability in a coin flip model with 4 equally likely outcomes. Given event B, we restrict attention to at least one H: HH, HT, TH, which are 3 outcomes. Among them, the matching flips in A are only HH, so 1 outcome. The fraction 1/3 is P(A|B), showing the share of at-least-one-H outcomes where flips match. A common mistake is using all 4 outcomes as the denominator, like 2/4 for all matches, ignoring the condition. A helpful strategy is to circle B outcomes first, then count A inside to focus on the restricted space.
A card is drawn at random from a set of 12 equally likely cards labeled:
{S1, S2, S3, S4, H1, H2, H3, H4, D1, D2, D3, D4}
The letter indicates the suit (S, H, D) and the number is 1–4.
Let event A = “the suit is H” and event B = “the number is 3 or 4.”
Using the outcomes listed, what fraction of outcomes in B also belong to A? Answer as a fraction.
Explanation: This question tests conditional probability using a model with listed outcomes. 'Given B' means we restrict our attention to only the outcomes in B, which are cards with number 3 or 4. There are 6 outcomes in B: S3, S4, H3, H4, D3, D4. Of these, 2 outcomes are also in A, meaning the suit is H: H3, H4. The fraction 2/6 or 1/3 represents the proportion of B's outcomes that satisfy A. A common mistake is using the total number of H suits as numerator over total cards, like 4/12=1/3, but that's P(A); or wrong denominator for 2/4=1/2. A good strategy is to first circle the outcomes in B, then count how many of those are in A.
A student randomly selects one of the 8 equally likely outcomes shown in the table.
Let event A = “the outcome has a star” and event B = “the outcome is in Row 2.”
Using the table, what is P(A∣B)? Give your answer as a fraction.
Table of outcomes:
Explanation: This question tests conditional probability using a model with listed outcomes in a table. 'Given B' means we restrict our attention to only the outcomes in B, which are in Row 2. There are 4 outcomes in B: 5, 6★, 7, 8★. Of these, 2 outcomes are also in A, meaning they have a star: 6★ and 8★. The fraction 2/4 or 1/2 represents P(A|B) because it shows the proportion of B's outcomes that satisfy A. A common mistake is using the total outcomes as the denominator, like 4 stars out of 8 for 1/2, but that's P(A); or counting only one star in Row 2 wrongly. A good strategy is to first circle the outcomes in B, then count how many of those are in A.
A two-stage experiment is performed: first choose a shape from {Circle, Square}, then choose a color from {Red, Blue, Green}. All outcomes are equally likely.
Outcomes: (Circle,Red) (Circle,Blue) (Circle,Green) (Square,Red) (Square,Blue) (Square,Green)
Let event A = “the shape is Square” and event B = “the color is Blue or Green.”
From the outcomes listed, what fraction of outcomes in B also belong to A? Answer as a fraction.
Explanation: This question tests conditional probability using a model with listed outcomes. 'Given B' means we restrict our attention to only the outcomes in B, which are those with color Blue or Green. There are 4 outcomes in B: (Circle,Blue), (Circle,Green), (Square,Blue), (Square,Green). Of these, 2 outcomes are also in A, meaning the shape is Square: (Square,Blue) and (Square,Green). The fraction 2/4 or 1/2 represents the proportion of B's outcomes that satisfy A. A common mistake is using the total outcomes as the denominator, like 3 squares out of 6 for 1/2, but that's P(A); or counting only one color wrongly for 1/3. A good strategy is to first circle the outcomes in B, then count how many of those are in A.
A two-stage experiment is performed: pick one letter from {X, Y, Z}, then pick one number from {1, 2}. All outcomes are equally likely.
Outcomes: (X,1) (X,2) (Y,1) (Y,2) (Z,1) (Z,2)
Let event A = “the letter is Z” and event B = “the number is 2.”
Based on the outcomes listed, what is P(A∣B)? Give your answer as a fraction.
Explanation: This question tests conditional probability using a model with listed outcomes. 'Given B' means we restrict our attention to only the outcomes in B, which are those with the number 2. There are 3 outcomes in B: (X,2), (Y,2), (Z,2). Of these, 1 outcome is also in A, meaning the letter is Z: (Z,2). The fraction 1/3 represents P(A|B) because it shows the proportion of B's outcomes that satisfy A. A common mistake is using the total number of outcomes as the denominator, like 2 Z out of 6 total for 1/3, but that's P(A); or counting wrong to get 1/2. A good strategy is to first circle the outcomes in B, then count how many of those are in A.
A lunch combo is chosen at random from the 12 equally likely combinations formed by choosing 1 sandwich and 1 drink:
Sandwiches: Turkey (T), Veggie (V), Ham (H) Drinks: Water (W), Juice (J), Soda (S), Tea (Te)
The 12 outcomes are: TW, TJ, TS, TTe, VW, VJ, VS, VTe, HW, HJ, HS, HTe. Let event A be “the drink is Juice,” and let event B be “the sandwich is Ham.” Using the outcomes listed, what is P(A∣B)? Give your answer as a fraction.
Explanation: This question tests conditional probability using a list of lunch combo outcomes. 'Given B' means we restrict our attention only to the combos with a Ham sandwich. There are 4 such outcomes (HW, HJ, HS, HTe). Among these 4, 1 has Juice (HJ). Therefore, the fraction 1/4 represents the conditional probability of Juice given Ham. A common mistake is using the total 12 combos as the denominator, like 1/12, but that ignores the sandwich condition. A good strategy is to first circle or list the outcomes in B, then count how many of those are also in A.
A jar contains 12 equally likely marbles: 4 red (R) and 8 blue (B). Each marble is also either striped (S) or plain (P) as shown by the list of outcomes (color, pattern):
(R,S), (R,S), (R,P), (R,P), (B,S), (B,S), (B,S), (B,P), (B,P), (B,P), (B,P), (B,P)
Let event A be “the marble is striped,” and let event B be “the marble is blue.” Among outcomes in B, what fraction also belong to A? Give your answer as a fraction.
Explanation: This question tests conditional probability using a list of marble outcomes. 'Given B' means we restrict our attention only to the blue marbles. There are 8 blue marbles in total. Among these 8, 3 are striped. Therefore, the fraction 3/8 represents the conditional probability of drawing a striped marble given that it is blue. A common mistake is using the total 12 marbles as the denominator, getting 3/12, but that ignores the condition of being blue. A good strategy is to first circle or list the outcomes in B, then count how many of those are also in A.
A box contains 10 equally likely balls. Each ball is labeled with a color and a number. The outcomes are:
R1, R2, R3, R4, R5, B1, B2, B3, B4, B5
Let event A be “the number is odd,” and let event B be “the ball is blue.” Based on the outcomes listed, what fraction of outcomes in B also belong to A? Give your answer as a fraction.
Explanation: This question tests conditional probability using a list of ball outcomes. 'Given B' means we restrict our attention only to the blue balls. There are 5 blue balls (B1 through B5). Among these 5, 3 have odd numbers (B1, B3, B5). Therefore, the fraction 3/5 represents the conditional probability of an odd number given that the ball is blue. A common mistake is using the total 10 balls as the denominator, like 3/10, but that overlooks the color condition. A good strategy is to first circle or list the outcomes in B, then count how many of those are also in A.
A two-stage experiment is performed: flip a fair coin (H or T), then roll a fair number cube (1–6). The 12 equally likely outcomes are:
H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6
Let event A be “the roll is even,” and let event B be “the coin lands T.” From the outcomes listed, what fraction of outcomes in B also belong to A? Give your answer as a fraction.
Explanation: This question tests conditional probability using a list of experiment outcomes. 'Given B' means we restrict our attention only to the outcomes where the coin lands T. There are 6 such outcomes (T1 through T6). Among these 6, 3 have an even roll (T2, T4, T6). Therefore, the fraction 3/6 represents the conditional probability of an even roll given tails. A common mistake is using the total 12 outcomes as the denominator, like 3/12, but that doesn't focus on the tails condition. A good strategy is to first circle or list the outcomes in B, then count how many of those are also in A.
A bag contains 9 equally likely tokens, each showing a letter and a shape:
(A, circle), (A, circle), (A, square), (B, circle), (B, square), (B, square), (C, circle), (C, square), (C, square)
Let event A be “the token shows a square,” and let event B be “the token shows the letter C.” Using the outcomes listed, what is P(A∣B)? Give your answer as a fraction.
Explanation: This question tests conditional probability using a list of token outcomes. 'Given B' means we restrict our attention only to the tokens showing the letter C. There are 3 such tokens (C circle, C square, C square). Among these 3, 2 show a square. Therefore, the fraction 2/3 represents the conditional probability of a square given that the token shows C. A common mistake is using the total 9 tokens as the denominator, like 2/9, but that ignores the condition of letter C. A good strategy is to first circle or list the outcomes in B, then count how many of those are also in A.
A spinner has 8 equally likely outcomes labeled 1 through 8. The outcomes are grouped below:
Let event A be “the outcome is prime,” and let event B be “the outcome is odd.” Based on the outcomes listed, what fraction of outcomes in B also belong to A? Give your answer as a fraction.
Explanation: This question tests conditional probability using a list of spinner outcomes. 'Given B' means we restrict our attention only to the odd outcomes. There are 4 odd outcomes (1, 3, 5, 7). Among these 4, 3 are prime (3, 5, 7). Therefore, the fraction 3/4 represents the conditional probability of a prime outcome given that it is odd. A common mistake is using the total 8 outcomes as the denominator, like 3/8, but that overlooks the restriction to odd outcomes. A good strategy is to first circle or list the outcomes in B, then count how many of those are also in A.
A spinner has 8 equal sections labeled 1 through 8. One spin is made. The outcomes are equally likely.
Let event A be “the result is even,” and let event B be “the result is greater than 4.”
What is P(A∣B)? Write your answer as a simplified fraction.
Explanation: This question tests conditional probability from a spinner model. The phrase "given B" means we restrict our attention to only the outcomes in event B. Event B is "the result is greater than 4," which includes the outcomes {5, 6, 7, 8}—that's 4 outcomes. Among these 4 outcomes in B, we need to count how many are also in event A ("the result is even"). The even numbers in {5, 6, 7, 8} are 6 and 8, giving us 2 outcomes. Therefore, P(A|B) = 2/4 = 1/2. A common mistake is using 8 (all outcomes) as the denominator instead of 4 (outcomes in B only)—remember to first circle the outcomes in B, then count how many of those are also in A.
A student randomly selects one of the 9 equally likely ordered pairs shown:
{(R,1),(R,2),(R,3),(G,1),(G,2),(G,3),(B,1),(B,2),(B,3)}
Let event A be “the color is G,” and let event B be “the number is 2 or 3.”
What is P(A∣B)? Write your answer as a simplified fraction.
Explanation: This question tests conditional probability from an ordered pair model. When finding P(A|B), we restrict our attention to outcomes in event B. Event B is "the number is 2 or 3," which includes pairs {(R,2), (R,3), (G,2), (G,3), (B,2), (B,3)}—that's 6 outcomes. Among these 6 pairs with number 2 or 3, we need to count how many have color G (event A). The pairs (G,2) and (G,3) satisfy both conditions, giving us 2 outcomes. Therefore, P(A|B) = 2/6 = 1/3. A common error is using 9 (all pairs) in the denominator—always remember that "given B" means B becomes your new sample space.
A card is drawn from a set of 12 equally likely cards labeled A1,A2,A3,A4,B1,B2,B3,B4,C1,C2,C3,C4.
Let event A be “the letter is B,” and let event B be “the number is even.”
What is P(A∣B)? Write your answer as a simplified fraction.
Explanation: This question tests conditional probability from a card-drawing model. The notation P(A|B) means we restrict attention to outcomes in event B. Event B is "the number is even," which includes cards {A2, A4, B2, B4, C2, C4}—that's 6 outcomes. Among these 6 cards with even numbers, we need to count how many have the letter B (event A). The cards B2 and B4 satisfy both conditions, giving us 2 outcomes. Therefore, P(A|B) = 2/6 = 1/3. A common error is using 12 (all cards) as the denominator instead of 6 (even-numbered cards only)—always start by identifying which outcomes are in the given condition B.
A box contains 12 equally likely slips: 5 say “Math,” 4 say “Science,” and 3 say “History.” One slip is chosen.
Let event A be “the slip says Science,” and let event B be “the slip does not say Math.”
What is P(A∣B)? Write your answer as a simplified fraction.
Explanation: This question tests conditional probability from a slip-drawing model. The notation P(A|B) means we consider only outcomes in event B. Event B is "the slip does not say Math," which includes Science and History slips: 4 + 3 = 7 slips total. Among these 7 non-Math slips, we need to count how many say Science (event A). There are 4 Science slips among the 7 non-Math slips. Therefore, P(A|B) = 4/7. A common mistake is dividing by 12 (all slips) instead of 7 (non-Math slips only)—remember that conditional probability restricts your sample space to the given condition.
Two fair coins are flipped. The equally likely outcomes are listed below:
{HH, HT, TH, TT}
Let event A be “exactly one head,” and let event B be “the first coin is heads.”
What fraction of outcomes in B also belong to A? Write your answer as a simplified fraction.
Explanation: This question tests conditional probability from a coin-flipping model. The key phrase "fraction of outcomes in B" tells us we're finding P(A|B). Event B is "the first coin is heads," which includes outcomes {HH, HT}—that's 2 outcomes. Among these 2 outcomes where the first coin is heads, we need to count how many have exactly one head total (event A). Only HT has exactly one head (HH has two heads), so that's 1 outcome. Therefore, the fraction is 1/2. A common mistake is counting all outcomes with exactly one head (HT and TH) without restricting to B first—remember to circle outcomes in B, then count A within that circle.
A password is formed by choosing one letter from {A,B,C,D} and then choosing one digit from {1,2,3}. All 12 outcomes are equally likely:
{A1,A2,A3,B1,B2,B3,C1,C2,C3,D1,D2,D3}
Event B: the digit is 1. Event A: the letter is C or D.
Based on the outcomes listed, which value best represents P(A∣B)? Give your answer as a simplified fraction.
Explanation: This question asks for P(A|B), the conditional probability that the letter is C or D given that the digit is 1. When we condition on B (digit is 1), we restrict our attention to only the passwords ending in 1. Event B consists of all outcomes with digit 1: {A1, B1, C1, D1}, giving us 4 outcomes in B. Among these 4 passwords ending in 1, we need to count how many have letters C or D (event A): C1 and D1 satisfy both conditions, so 2 outcomes are in both A and B. Therefore, P(A|B) = 2/4 = 1/2 when simplified. The key is to first identify all outcomes where the digit is 1, then count within that restricted set.