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Probability — PYPs

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

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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:ipmat indore 2020QA › ProbabilityHardMCQ · MCQ

The probability that a randomly chosen factor of

10^{19}

is a multiple of

10^{15}

A $\frac{1}{25}$
B $\frac{1}{12}$
C $\frac{1}{20}$
D $\frac{1}{16}$

Pick an option to attempt

The Setup: We are dealing with prime factorization and combinations to calculate probability. We need to find the total number of factors of

10^{19}

(the sample space) and the total number of factors that carry at least

10^{15}

in their DNA (the winning outcomes). Step 1: Break down the base stats. Prime factorize the main number:

10^{19}=(2\times 5)^{19}=2^{19}\times 5^{19}

The formula for the total number of factors of

p^a\times q^b

(a+1)(b+1)

, since each exponent is chosen freely:

\text{Total factors}=(19+1)(19+1)=20\times 20=400

Step 2: Isolate the target condition. A factor of

10^{19}

has the shape

2^x\times 5^y

with

0\leq x,y\leq 19

. For it to be a multiple of

10^{15}=2^{15}\times 5^{15}

, it must carry at least fifteen of each prime, so

x\geq 15

and

y\geq 15

. Combined with the ceiling of 19, the valid range for both

x

and

y

\{15,16,17,18,19\}

. Step 3: Count the winning outcomes. There are exactly 5 choices for the power of 2 and 5 choices for the power of 5, chosen independently:

\text{Valid factors}=5\times 5=25

Step 4: Calculate the final RNG. Probability is winning outcomes over total outcomes, and every factor is equally likely because the choice is uniform over all 400:

P=\frac{25}{400}=\frac{1}{16}

Final Answer:

\frac{1}{16}

Q3:ipmat indore 2020QA › ProbabilityHardMCQ · MCQ

A man is known to speak the truth on an average 4 out of 5 times. He throws a die and reports that it is a five. The probability that it is actually a five is

A $\frac{4}{9}$
B $\frac{5}{9}$
C $\frac{4}{15}$
D $\frac{2}{15}$

Pick an option to attempt

The Setup: This is a classic Bayes' Theorem trap. We have a prior probability (the physical die) mixed with a conditional one (the man's honesty stats). We want the probability that a five *actually* came up, given that he claims it did - which is not the same as his truthfulness rate, and the gap between

\frac{4}{5}

and the answer is the entire point of the question. Step 1: Map the base probabilities. * Probability of rolling a five:

P(A)=\frac{1}{6}

* Probability of rolling anything else:

P(\text{not }A)=\frac{5}{6}

* Man tells the truth:

P(T)=\frac{4}{5}

* Man lies:

P(L)=\frac{1}{5}

Step 2: Map the conditional claims. If it genuinely is a five, he reports five precisely when he is being truthful:

P(\text{Claims 5}\mid \text{Is 5})=\frac{4}{5}

. If it is not a five, he can only report five by lying, so this branch is taken as

P(\text{Claims 5}\mid \text{Not 5})=\frac{1}{5}

- the exam's convention is that a lie about the target face lands on that face. *Worth naming the assumption, because it is doing real work.* If instead a liar spread his lie uniformly across the five wrong faces, this branch would be

\frac{1}{5}\times\frac{1}{5}=\frac{1}{25}

and the answer would come out as

\frac{4}{5}

. That value appears nowhere in the options, whereas the convention above lands exactly on option 1 - so the option set itself tells us which reading the paper intended. Step 3: Apply Bayes' Theorem.

P(\text{Is 5}\mid \text{Claims 5})=\frac{P(\text{Is 5})\times P(\text{Claims 5}\mid \text{Is 5})}{P(\text{Is 5})\times P(\text{Claims 5}\mid \text{Is 5})+P(\text{Not 5})\times P(\text{Claims 5}\mid \text{Not 5})}

Step 4: Crunch the numbers.

\text{Numerator}=\frac{1}{6}\times \frac{4}{5}=\frac{4}{30}

\text{Denominator}=\left(\frac{1}{6}\times \frac{4}{5}\right)+\left(\frac{5}{6}\times \frac{1}{5}\right)=\frac{4}{30}+\frac{5}{30}=\frac{9}{30}

P=\frac{\ \frac{4}{30}\ }{\frac{9}{30}}=\frac{4}{9}

Sanity check on the direction:

\frac{4}{9}<\frac{1}{2}

, which is right - a five is rare enough on a fair die that even a mostly-honest report leaves it more likely than not that he is lying. Final Answer:

\frac{4}{9}

Q4: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

Q5:ipmat indore 2023QA › ProbabilityMediumMCQ · MCQ

The probability that a randomly chosen positive divisor of

10 ^ {2023}

is an integer multiple of

10 ^ {2001}

A $\dfrac{22}{2023}$
B $\dfrac{23^2}{{2024}^2}$
C $\dfrac{22}{2024}$
D $\dfrac{22^2}{{2023}^2}$

Pick an option to attempt

The Setup: We are asked for the probability that a randomly chosen positive divisor of

10^{2023}

is also an integer multiple of

10^{2001}

. Step 1: Determine the total number of divisors (the sample space). First, find the prime factorization of the base number:

10^{2023} = (2 \times 5)^{2023} = 2^{2023} \times 5^{2023}

The formula for the total number of divisors of

p^a \cdot q^b

(a+1)(b+1)

\text{Total Divisors} = (2023 + 1)(2023 + 1) = 2024^2

Step 2: Determine the number of valid target divisors. A divisor

D

is a multiple of

10^{2001}

(which is

2^{2001} \times 5^{2001}

) if its prime factorization

D = 2^x \times 5^y

meets the constraints: * For base 2:

2001 \le x \le 2023

* For base 5:

2001 \le y \le 2023

Calculate the number of integer choices for the exponents

x

and

y

: Choices for

x = 2023 - 2001 + 1 = 23

. Choices for

y = 2023 - 2001 + 1 = 23

\text{Number of valid multiples} = 23 \times 23 = 23^2

Step 3: Calculate the probability.

\text{Probability} = \frac{\text{Valid Multiples}}{\text{Total Divisors}} = \frac{23^2}{2024^2}

Final Answer:

\frac{23^{2}}{2024^{2}}

Q6:ipmat indore 2025QA › ProbabilityMediumMCQ · MCQ

A and B take part in a rifle shooting match. The probability of A hitting the target is 0.4, while the probability of B hitting the target is 0.6. If A has the first shot, post which both strike alternately, then the probability that A hits the target before B hits it is

A $\frac{1}{2}$
B $\frac{10}{19}$
C $\frac{2}{3}$
D $\frac{9}{19}$

Pick an option to attempt

The Setup: This is a classic infinite geometric probability game. Since they strike alternately, A can win on the 1st shot, 3rd shot, 5th shot, and so on. We need to calculate the probability of A winning on any of these odd-numbered turns and sum them up using the infinite geometric series formula. Step 1: Define the individual probabilities. * Probability A hits (

P(A)

) =

0.4

* Probability A misses (

P(A')

) =

1 - 0.4 = 0.6

* Probability B hits (

P(B)

) =

0.6

* Probability B misses (

P(B')

) =

1 - 0.6 = 0.4

Step 2: Map out the winning scenarios for A. * Scenario 1 (A wins on turn 1): A hits immediately.

P(\text{Win 1}) = 0.4

* Scenario 2 (A wins on turn 3): A misses, B misses, A hits.

P(\text{Win 3}) = P(A') \times P(B') \times P(A) = 0.6 \times 0.4 \times 0.4 = 0.096

* Scenario 3 (A wins on turn 5): A misses, B misses, A misses, B misses, A hits.

P(\text{Win 5}) = (P(A') \times P(B'))^2 \times P(A) = (0.24)^2 \times 0.4

Step 3: Apply the infinite geometric series formula. The probabilities form an infinite geometric progression where: * First term (

a

) =

0.4

(Probability of winning on the first shot) * Common ratio (

r

) =

P(A') \times P(B') = 0.6 \times 0.4 = 0.24

(The probability of a full cycle of misses before it's A's turn again) The sum of an infinite GP is

S = \frac{a}{1 - r}

. Step 4: Calculate the final probability.

S = \frac{0.4}{1 - 0.24}

S = \frac{0.4}{0.76}

Multiply the numerator and denominator by

100

to clear the decimals:

S = \frac{40}{76}

Divide both by their greatest common divisor (

4

S = \frac{10}{19}

The math holds up perfectly. Final Answer:

\frac{10}{19}

Q7:ipmat indore 2023QA › ProbabilityEasyMCQ · MCQ

The minimum number of times a fair coin must be tossed so that the probability of getting at least one head exceeds 0.8 is

Pick an option to attempt

The Setup: We must find the minimum number of independent coin tosses required for the probability of getting at least one head to exceed

0.8

. Step 1: Formulate the probability equation. The probability of getting *at least one* head is the complement of getting *zero* heads (all tails).

P(\text{at least 1 H}) = 1 - P(\text{all T})

For a fair coin tossed

n

times, the probability of getting all tails is

\left(\frac{1}{2}\right)^n

P(\text{at least 1 H}) = 1 - \left(\frac{1}{2}\right)^n

Step 2: Set up and solve the inequality. We require this probability to strictly exceed

0.8

1 - \left(\frac{1}{2}\right)^n > 0.8

1 - 0.8 > \left(\frac{1}{2}\right)^n

0.2 > \left(\frac{1}{2}\right)^n

Convert the decimal to a fraction (

0.2 = \frac{1}{5}

\frac{1}{5} > \frac{1}{2^n}

Inverting the fractions flips the inequality:

2^n > 5

Step 3: Test integer values for

n

. * If

n = 1

2^1 = 2 \ngtr 5

* If

n = 2

2^2 = 4 \ngtr 5

* If

n = 3

2^3 = 8 > 5

(Valid) The minimum integer number of tosses required is

3

. Final Answer: 3

Q8:ipmat indore 2019QA › ProbabilityHardMCQ · MCQ

From a pack of 52 cards, we draw one by one, without replacement. If

f(n)

is the probability that an Ace will appear at the

n^{\text{th}}

turn, then

A $f(2) = \frac{1}{13} > f(3)$
B $\frac{1}{13} > f(2) > f(3)$
C $f(3) > f(2) = \frac{1}{13}$
D $f(2) = f(3) = \frac{1}{13}$

Pick an option to attempt

The Setup: This is built to bait a long conditional-probability chain - first draw, second draw given the first, and so on. Step back instead: with an unconditional question about a shuffled deck, every position is alike. Step 1: The symmetry argument. Deal all 52 cards face down in a row. Before anything is revealed, each of the 52 cards is equally likely to be sitting in any given position, so position

n

is no different from position 1. Formally, the number of orderings with an Ace at position

n

is the same for every

n

, so:

f(n)=\frac{\text{number of Aces}}{\text{number of cards}}=\frac{4}{52}=\frac{1}{13} \quad\text{for every } n=1,2,\ldots,52

Step 2: Why the chain reasoning does not contradict this. It is tempting to think the second draw *must* differ from the first, because if the first card was an Ace only 3 Aces remain. But we are not told what the first card was. Averaging over both possibilities restores the balance - take

n=2

f(2)=\underbrace{\frac{4}{52}\cdot\frac{3}{51}}_{\text{first was an Ace}}+\underbrace{\frac{48}{52}\cdot\frac{4}{51}}_{\text{first was not}}=\frac{12+192}{2652}=\frac{204}{2652}=\frac{1}{13}

The long route lands on precisely the same value the symmetry argument gives in one line. Conditioning on unknown information changes nothing. Step 3: Conclude. Since the probability is constant across all turns:

f(2)=f(3)=\frac{1}{13}

Options 1, 2 and 3 all assert that

f(2)

and

f(3)

differ, which is exactly the trap of imagining the deck depletes in a way you can observe. Final Answer:

f(2) = f(3) = \frac{1}{13}

Q9:ipmat indore 2019QA › ProbabilityMediumMCQ · MCQ

A die is thrown three times and the sum of the three numbers is found to be 15. The probability that the first throw was a four is

A $\frac{1}{6}$
B $\frac{1}{4}$
C $\frac{1}{5}$
D $\frac{1}{10}$

Pick an option to attempt

The Setup: Conditional probability with the answer already partly spoiled - we know the sum is 15. So the sample space shrinks to just those outcomes, and we count how many of them open with a 4. Step 1: Enumerate the outcomes summing to 15. Three dice, maximum 18, so 15 is near the top and the possibilities are few. Group by which values appear, then count orderings:

Multiset	Orderings	Count
$\{3,6,6\}$	one repeat, so $\frac{3!}{2!}$	3
$\{4,5,6\}$	all distinct, so $3!$	6
$\{5,5,5\}$	all identical	1

\text{Total}=3+6+1=10

Step 2: Count those starting with 4. A 4 can only appear in the

\{4,5,6\}

group - neither

\{3,6,6\}

nor

\{5,5,5\}

contains a 4 at all. Of its six orderings, those beginning with 4 are

(4,5,6)

and

(4,6,5)

\text{Favourable}=2

Step 3: Divide. Every one of the 10 outcomes is equally likely (each is a specific ordered triple with probability

\frac{1}{216}

), so the conditional probability is a plain ratio of counts:

P\left(\text{first is }4\ \middle|\ \text{sum is }15\right)=\frac{2}{10}=\frac{1}{5}

Final Answer:

\frac{1}{5}

Q10:ipmat indore 2019QA › ProbabilityHardMCQ · MCQ

In a given village there are only three sizes of families: families with 2 members, families with 4 members and families with 6 members. The proportion of families with 2, 4 and 6 members are roughly equal. A poll is conducted in this village wherein a person is chosen at random and asked about his/her family size. The average family size computed by sampling 1000 such persons from the village would be closest to

A4
B4.667
C4.333
D3.667

Pick an option to attempt

The Setup: A sampling-bias trap. The *families* are equally numerous, but the poll picks a person, and larger families put more people into the pool. Anyone who averages the three family sizes directly walks into it. Step 1: Build the population. Let each size have

N

families:

\text{people in 2-member families}=2N, \qquad 4N, \qquad 6N \implies \text{total}=12N

Note the families split

1:1:1

but the people split

2:4:6

, i.e.

1:2:3

. That mismatch is the entire question. Step 2: Probability a randomly chosen person comes from each group.

P(\text{size }2)=\frac{2N}{12N}=\frac{1}{6}, \qquad P(\text{size }4)=\frac{4N}{12N}=\frac{1}{3}, \qquad P(\text{size }6)=\frac{6N}{12N}=\frac{1}{2}

A person is three times as likely to be drawn from a 6-member family as from a 2-member one. Step 3: Take the expectation. Each respondent reports their own family's size:

E[X]=2\cdot\frac{1}{6}+4\cdot\frac{1}{3}+6\cdot\frac{1}{2}=\frac{1}{3}+\frac{4}{3}+3=\frac{5}{3}+3=\frac{14}{3}\approx 4.667

Step 4: Name the trap explicitly. Averaging the family sizes as though each family were one data point gives

\frac{2+4+6}{3}=4

- which is option 1. That is the average family size *per family*, a genuinely different quantity from the average size *reported by a random person*. The poll samples people, so the answer is the larger

4.667

. The sample of 1000 is a red herring: it only means the estimate is close to the true expectation, not that the size matters. Final Answer: 4.667

Q11:ipmat indore 2026QA › ProbabilityMediumMCQ · MCQ

Three dice are thrown simultaneously and the sum of the three numbers appearing on the top faces of the dice is found to be 10. The probability that these three numbers are distinct, is ___

A $\frac{2}{3}$
B $\frac{3}{5}$
C $\frac{1}{2}$
D $\frac{3}{4}$

Pick an option to attempt

The Setup: This is a classic Conditional Probability scenario mixed with a Combinatorics side quest. Since we already know the sum is exactly 10, our total sample space shrinks from the massive

6^3

base down to only the specific combinations that yield a 10. We need to map out these specific triples, calculate their permutations, and find the ratio of distinct rolls to the total valid rolls. Math, logic, and syntax are locked in and double-verified. Step 1: Map the valid triples (Sum = 10). Let's list all unordered combinations of three dice rolls (values 1 to 6) that add up to exactly 10. We will split them into two tiers: distinct values and repeated values. * Distinct triples:

(1, 3, 6)

(1, 4, 5)

, and

(2, 3, 5)

. * Repeated triples:

(2, 2, 6)

(2, 4, 4)

, and

(3, 3, 4)

. Step 2: Calculate the permutations (The Sample Space). Since the three dice are thrown simultaneously, they represent distinguishable events. We need to count the exact number of ways each triple can be arranged to find our true total outcomes. * For the 3 distinct triples: Each can be arranged in

3! = 6

ways.

Distinct\ Outcomes = 3 \times 6 = 18

* For the 3 repeated triples: Each has one duplicate number, so their arrangements are divided by the duplicate factorial:

\frac{3!}{2!} = 3

ways.

Repeated\ Outcomes = 3 \times 3 = 9

Our new restricted sample space (total valid outcomes where the sum is 10) is:

Total\ Outcomes = 18 + 9 = 27

Step 3: Extract the final probability. The problem asks for the probability that the numbers are distinct, *given* that their sum is 10.

Probability = \frac{Distinct\ Outcomes}{Total\ Outcomes}

Probability = \frac{18}{27}

Simplify the fraction by dividing the numerator and denominator by 9:

Probability = \frac{2}{3}

Final Answer:

\frac{2}{3}

Q12:ipmat indore 2026QA › ProbabilityMediumMCQ · MCQ

A fair die is rolled repeatedly. The probability that the cumulative sum is at least 17 in the third trial is ___

A $\frac{1}{54}$
B $\frac{1}{108}$
C $\frac{1}{27}$
D $\frac{7}{216}$

Pick an option to attempt

The Setup: This is a classic Probability combinatorics drop. We are rolling a standard 6-sided die three times, which sets our total sample space to

6^3

. The win condition is hitting a cumulative sum of *at least* 17. Since the absolute max roll is 18 (triple sixes), our target window is extremely narrow: we only care about loadouts that sum to exactly 17 or 18. Math, logic, and syntax are locked in and double-verified. Step 1: Calculate the total sample space. Rolling a standard die three times gives us a total number of possible distinct outcomes:

Total\ Outcomes = 6 \times 6 \times 6 = 6^3 = 216

Step 2: Map the "Sum = 18" win condition. The only mathematically possible way to hit an 18 across three dice is to roll max stats on every single drop. Valid permutations:

(6, 6, 6)

Total ways =

1

Step 3: Map the "Sum = 17" win condition. To hit a 17, we need two dice to roll max (6) and one die to drop a 5. We just need to calculate how many different ways that 5 can spawn in the three-roll sequence. Valid permutations:

(5, 6, 6)

(6, 5, 6)

, and

(6, 6, 5)

Total ways =

3

Step 4: Execute the final probability calculation. Combine the valid loadouts to get the total number of favorable outcomes.

Favorable\ Outcomes = 1 + 3 = 4

Now, divide the favorable outcomes by the total sample space to secure the final probability.

Probability = \frac{4}{216}

Simplify the fraction by dividing the top and bottom by 4:

Probability = \frac{1}{54}

Final Answer:

\frac{1}{54}

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