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Probability Interview Questions

Probability questions, with multiple-choice options. The step-by-step answers are at the bottom.

Regular Probability Questions

  1. Dice Roll: What is the probability of rolling a number greater than 4 on a standard six-sided die?

    A) 1/6

    B) 1/3

    C) 1/2

    D) 2/3

  1. Coin Flips: If you flip three fair coins, what is the probability of getting exactly two heads?

    A) 1/8

    B) 1/2

    C) 2/3

    D) 3/8

  1. Marbles in a Jar: A jar contains 5 red marbles, 3 blue marbles, and 2 green marbles. If you draw one marble at random, what is the probability that it is not blue?

    A) 3/10

    B) 1/2

    C) 7/10

    D) 8/10

  1. Standard Deck of Cards: What is the probability of drawing a face card (Jack, Queen, or King) from a standard 52-card deck?

    A) 1/13

    B) 3/52

    C) 3/13

    D) 1/4

  1. Union of Events: From a standard 52-card deck, what is the probability of drawing a card that is either a spade or a Queen?

    A) 17/52

    B) 1/4

    C) 3/13

    D) 4/13

Conditional Probability Questions

  1. Two Dice: You roll two standard six-sided dice. Given that the sum of the two dice is 8, what is the probability that one of the dice shows a 5?

    A) 1/6

    B) 1/5

    C) 2/5

    D) 1/3

  1. Drawing Cards: You draw two cards from a standard 52-card deck without replacement. What is the probability that the second card is a King, given that the first card was also a King?

    A) 3/52

    B) 1/26

    C) 1/17

    D) 1/13

  1. School Survey: At a school, 40% of students play basketball, and 25% of students play soccer. Of the students who play basketball, 30% also play soccer. What is the probability that a randomly selected student plays both basketball and soccer?

    A) 10%

    B) 12%

    C) 30%

    D) 70%

  1. Medical Test: A certain disease is present in 1 out of 1000 people. A test for the disease is 99% accurate (i.e., it gives a positive result for 99% of people who have the disease and a negative result for 99% of people who don't). If a person tests positive, what is the probability they actually have the disease?

    A) 0.1%

    B) 1%

    C) 9%

    D) 99%

  1. Two Marbles: A bag contains 4 red marbles and 6 blue marbles. You randomly draw two marbles from the bag without replacement. Given that the second marble you draw is blue, what is the probability that the first marble you drew was red?

    A) 2/5

    B) 1/2

    C) 4/9

    D) 2/3

Step-by-Step Answers

Regular Probability Answers

  1. (B) Dice Roll: The favorable outcomes are {5, 6}. The total outcomes are {1, 2, 3, 4, 5, 6}. So .

  2. (D) Coin Flips: The sample space has 8 outcomes {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}. The favorable outcomes (exactly two heads) are {HHT, HTH, THH}. So .

  3. (C) Marbles in a Jar: There are 10 marbles total. The number of non-blue marbles is 5 (red) + 2 (green) = 7. So P=7/10

  4. (C) Standard Deck of Cards: There are 3 face cards per suit and 4 suits, so 3×4=12 face cards. So P=12/52=3/13.

  5. (D) Union of Events: Use the formula P(A∪B)=P(A)+P(B)−P(A∩B).

    P(Spade or Queen)=P(Spade)+P(Queen)−P(Queen of Spades)

    =5213​+524​−521​=5216​=134​.

Conditional Probability Answers

  1. (C) Two Dice: The possible outcomes that sum to 8 are {(2,6), (3,5), (4,4), (5,3), (6,2)}. This is our reduced sample space of 5 outcomes. Of these, the outcomes with a 5 are {(3,5), (5,3)}. There are 2 such outcomes. So P=2/5

  2. (C) Drawing Cards: After one King is drawn, there are 51 cards left in the deck. Of those remaining cards, 3 are Kings. So, the probability is 3/51=1/17

  3. (B) School Survey: We want P(Soccer and Basketball). The formula is P(S∩B)=P(S∣B)×P(B).

    P(S∩B)=0.30×0.40=0.12, or 12%.

  4. (C) Medical Test: Using Bayes' Theorem: P(D∣+)=P(+)P(+∣D)P(D)​.

    The numerator is (0.99)(0.001)=0.00099.

    The denominator is the total probability of testing positive: (0.99)(0.001)+(0.01)(0.999)=0.00099+0.00999=0.01098.

    P(D∣+)=0.010980.00099​≈0.09016, which is about 9%.

  5. (C) Two Marbles: Using the formula P(R1​∣B2​)=P(B2​)P(R1​∩B2​)​.

    P(R1​∩B2​)=104​×96​=9024​.

    P(B2​)=P(R1​∩B2​)+P(B1​∩B2​)=(104​×96​)+(106​×95​)=9024​+9030​=9054​.

    P(R1​∣B2​)=54/9024/90​=5424​=94​.

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