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Probability questions for IPMAT, with solutions

10 free Probability practice questions for IPMAT Indore, Rohtak and JIPMAT, 5 from past papers, each with the answer and a full worked solution. No sign-in. Try each one first, then open the solution. The lessons for this chapter are in Probability on Learn.

  1. Question 1Multiple choiceAsked in JIPMAT 2023Classical Probability

    Given below are two statements in the context of throwing two different dice together:

    Statement (I): The probability that the two numbers obtained have even sum, is 0.500.50.

    Statement (II): The probability that the two numbers obtained have even product is 0.750.75.

    In light of the above statements, choose the most appropriate answer from the options given below.

    1. A.Both Statement (I) and Statement (II) are true.
    2. B.Both Statement (I) and Statement (II) are false.
    3. C.Statement (I) is true but Statement (II) is false.
    4. D.Statement (I) is false but Statement (II) is true.
    Show the answer and solution

    Answer: A. Both Statement (I) and Statement (II) are true.

    Each die is odd or even with probability 12\frac12, independently.

  2. Question 2Multiple choiceAsked in JIPMAT 2023Conditional Probability and Bayes

    Let AA and BB be two events such that P(A)=0.25P(A)=0.25, P(B)=0.50P(B)=0.50 and P(A∩B)=0.14P(A\cap B)=0.14. The probability that neither AA nor BB occurs is

    1. A.0.090.09
    2. B.0.190.19
    3. C.0.390.39
    4. D.0.290.29
    Show the answer and solution

    Answer: C. 0.390.39

    P(neither)=1−P(A∪B)=1−0.61=0.39P(\text{neither})=1-P(A\cup B)=1-0.61=0.39.

  3. Question 3Multiple choiceAsked in JIPMAT 2024Conditional Probability and Bayes

    A speaks the truth in 75%75\% cases and B in 80%80\% of the cases. In what percentage of cases are they likely to contradict each other in narrating the same incident?

    1. A.5%5\%
    2. B.15%15\%
    3. C.35%35\%
    4. D.45%45\%
    Show the answer and solution

    Answer: C. 35%35\%

    They contradict exactly when one tells the truth and the other does not.

  4. Question 4Multiple choiceWritten by IPMATwaySequential Events and Games

    A and B throw a fair die alternately, A first; the first to throw a six wins. What is the probability that A wins?

    1. A.511\frac{5}{11}
    2. B.16\frac{1}{6}
    3. C.611\frac{6}{11}
    4. D.12\frac{1}{2}
    Show the answer and solution

    Answer: C. 611\frac{6}{11}

    Each round with no six has probability 2536\frac{25}{36}.

  5. Question 5Multiple choiceAsked in IPMAT Indore 2023Classical Probability

    If a three-digit number is chosen at random, what is the probability that it is divisible neither by 33 nor by 44?

    1. A.14\frac{1}{4}
    2. B.23\frac{2}{3}
    3. C.12\frac{1}{2}
    4. D.13\frac{1}{3}
    Show the answer and solution

    Answer: C. 12\frac{1}{2}

    Density check: (1−13)(1−14)=12(1-\frac13)(1-\frac14)=\frac{1}{2}, and 900900 is a multiple of 1212, so the count is exact.

  6. Question 6Type the answerWritten by IPMATwayClassical Probability

    A bag contains nn balls, of which 33 are red. Two balls are drawn at random. If the probability that both are red is 115\frac{1}{15}, find nn.

    Show the answer and solution

    Answer: 10

    (32)=3\binom{3}{2}=3 favourable pairs; n(n−1)=90n(n-1)=90.

  7. Question 7Multiple choiceAsked in IPMAT Indore 2020Conditional Probability and Bayes

    A man is known to speak the truth on an average 44 out of 55 times. He throws a die and reports that it is a five. The probability that it is actually a five is

    1. A.49\frac{4}{9}
    2. B.59\frac{5}{9}
    3. C.415\frac{4}{15}
    4. D.215\frac{2}{15}
    Show the answer and solution

    Answer: A. 49\frac{4}{9}

    He reports 'five' when it is a five and he tells the truth (16⋅45=430\frac{1}{6}\cdot\frac{4}{5}=\frac{4}{30}), or when it is not a five and he lies (56⋅15=530\frac{5}{6}\cdot\frac{1}{5}=\frac{5}{30}) — the usual exam model treats any lie here as the report 'five'. Bayes: 44+5=49\frac{4}{4+5}=\frac{4}{9}.

  8. Question 8Multiple choiceWritten by IPMATwayProbability with Numbers

    An integer xx is chosen at random from 11 to 2020. What is the probability that x2−1x^{2}-1 is divisible by 33?

    1. A.710\frac{7}{10}
    2. B.13\frac{1}{3}
    3. C.310\frac{3}{10}
    4. D.23\frac{2}{3}
    Show the answer and solution

    Answer: A. 710\frac{7}{10}

    Among x−1,x,x+1x-1, x, x+1 one is a multiple of 33.

  9. Question 9Multiple choiceWritten by IPMATwaySequential Events and Games

    Two players play a best-of-three series (the first to win two games wins). A wins each game with probability 23\frac{2}{3}, independently. What is the probability that A wins the series?

    1. A.49\frac{4}{9}
    2. B.827\frac{8}{27}
    3. C.23\frac{2}{3}
    4. D.2027\frac{20}{27}
    Show the answer and solution

    Answer: D. 2027\frac{20}{27}

    A wins 22-00: (2/3)2=49(2/3)^2=\frac49. A wins 22-11: loses one of the first two games and wins the third: 2⋅23⋅13⋅23=8272\cdot\frac23\cdot\frac13\cdot\frac23=\frac{8}{27}. Total =1227+827=2027=\frac{12}{27}+\frac{8}{27}=\frac{20}{27}.

  10. Question 10Multiple choiceWritten by IPMATwaySequential Events and Games

    A marksman hits a target with probability 14\frac{1}{4} per shot, independently. He fires 44 shots. What is the probability that he hits the target at least twice?

    1. A.175256\frac{175}{256}
    2. B.27128\frac{27}{128}
    3. C.81256\frac{81}{256}
    4. D.67256\frac{67}{256}
    Show the answer and solution

    Answer: D. 67256\frac{67}{256}

    P(0)=(3/4)4=81256P(0)=(3/4)^4=\frac{81}{256}. P(1)=4⋅14⋅(3/4)3=108256P(1)=4\cdot\frac14\cdot(3/4)^3=\frac{108}{256}. So P(≥2)=1−189256=67256P(\ge2)=1-\frac{189}{256}=\frac{67}{256}.