JIPMAT Probability PYQs

2 solved Probability previous year questions (PYQs) from JIPMAT past year papers — attempt each and check the answer.

Q1:jipmat 2025QA › ProbabilityMediumQA · MCQ

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

A5%
B15%
C35%
D45%

Pick an option to attempt

The Setup: We are looking for the exact probability that one person is telling the truth while the other is straight capping (lying). Step 1: Define the probabilities. Let

P(A)

be the probability A tells the truth.

P(A)=0.75

. So, probability A lies is

P(A')=0.25

. Let

P(B)

be the probability B tells the truth.

P(B)=0.80

. So, probability B lies is

P(B')=0.20

. Step 2: Find the contradiction scenarios. They contradict when (A tells truth AND B lies) OR (A lies AND B tells truth).

P(\text{Contradict})=(P(A) \times P(B'))+(P(A') \times P(B))

Step 3: Plug in the numbers and do the math.

P(\text{Contradict})=(0.75 \times 0.20)+(0.25 \times 0.80)

P(\text{Contradict})=0.15+0.20=0.35

Step 4: Convert to a percentage.

0.35 \times 100\%=35\%

. Final Answer: 35%

Q2:jipmat 2025QA › ProbabilityMediumQA · MCQ

Four persons are chosen at random from a group of 3 men, 2 women and 4 children. The chance that exactly 2 of them are children is:

A $\frac{1}{9}$
B $\frac{2}{9}$
C $\frac{1}{12}$
D $\frac{10}{21}$

Pick an option to attempt

The Setup: We are building a squad of 4 people from a total pool of 9 (

3\text{M} + 2\text{W} + 4\text{C}

). To find the probability, we need to divide the number of 'winning' combinations (exactly 2 kids and 2 non-kids) by the total possible combinations. Step 1: Calculate the total number of ways to pick any 4 people out of 9. This is a standard combination formula

{}^nC_r = \frac{n!}{r!(n-r)!}

{}^9C_4 = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = \frac{3024}{24} = 126

So, there are

126

possible squads. Step 2: Calculate the favorable outcomes. We explicitly need *exactly*

2

children. That means the other

2

slots must be filled by adults (men or women). Total children =

4

. Ways to pick 2 kids =

{}^4C_2 = \frac{4 \times 3}{2 \times 1} = 6

. Total adults =

3 + 2 = 5

. Ways to pick 2 adults =

{}^5C_2 = \frac{5 \times 4}{2 \times 1} = 10

. Multiply them together to get the total favorable combos:

6 \times 10 = 60\text{ ways}

Step 3: Calculate the final probability.

P = \frac{\text{Favorable}}{\text{Total}} = \frac{60}{126}

Divide top and bottom by

6

P = \frac{10}{21}

Final Answer: 10/21

JIPMAT Probability — PYPs

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