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Home/JEE MAINS/mathematics/probability
Curated PYQ Collection

Top 50 Most Repeated PROBABILITY PYQs | JEE MAINS

A curated collection of the most important questions from PROBABILITY, fully solved with step-by-step concepts to prepare for JEE MAINS.

Question #1

Practice Question

A.$\frac{8}{20}$
B.$\frac{12}{20}$
C.$\frac{7}{20}$
D.$\frac{5}{20}$

Concept Applied

Total number of balls = 5 + 8 + 7 = 20. Number of non-white balls = 5 (red) + 7 (black) = 12. So, P(not white) = $\frac{12}{20}$. This simplifies to $...

Read Full Step-by-Step Solution →

Question #2

Practice Question

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

Concept Applied

After removing one red, 2 red and 2 green remain. P(green) = 2/4 = 1/2....

Read Full Step-by-Step Solution →

Question #3

Practice Question

A.$\frac{15}{64}$
B.$\frac{1}{4}$
C.$\frac{5}{16}$
D.$\frac{3}{8}$

Concept Applied

This is a binomial distribution with $n=6$, $p=0.5$, $k=4$. Use formula: $P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}$. So $P(X=4) = \binom{6}{4} (0.5)^4 (0...

Read Full Step-by-Step Solution →

Question #4

Practice Question

Concept Applied

Use P(A∪B) = P(A) + P(B) - P(A)P(B) for independent events....

Read Full Step-by-Step Solution →

Question #5

Practice Question

A.$\text{Mean: } \frac{5}{3}, \text{ Variance: } \frac{10}{9}$
B.$\text{Mean: } \frac{5}{3}, \text{ Variance: } \frac{20}{9}$
C.$\text{Mean: } \frac{5}{9}, \text{ Variance: } \frac{10}{3}$
D.$\text{Mean: } \frac{5}{9}, \text{ Variance: } \frac{20}{3}$

Concept Applied

Step 1: Recall the formula for the mean of a binomial distribution, which is $np$.,Step 2: Recall the formula for the variance of a binomial distribut...

Read Full Step-by-Step Solution →

Question #6

Practice Question

A.$\frac{1}{6}$
B.$\frac{1}{12}$
C.$\frac{1}{9}$
D.$\frac{5}{36}$

Concept Applied

Total outcomes = 36. Favorable outcomes for sum 7: (1,6),(2,5),(3,4),(4,3),(5,2),(6,1) → 6 cases. So, probability = $\frac{6}{36} = \frac{1}{6}$....

Read Full Step-by-Step Solution →

Question #7

Practice Question

A.\frac{1}{2}
B.\frac{1}{3}
C.\frac{2}{3}
D.\frac{5}{6}

Concept Applied

Even numbers on a die are {2,4,6} → P(A)=3/6=1/2. Complement probability =1−1/2=1/2....

Read Full Step-by-Step Solution →

Question #8

Practice Question

A.1/2
B.1/3
C.2/3
D.5/6

Concept Applied

Even numbers on a die are {2,4,6} (3 outcomes). The complement (odd numbers) also has 3 outcomes. Probability = 3/6 = 1/2....

Read Full Step-by-Step Solution →

Question #9

Practice Question

A.$\frac{1}{3}$
B.$\frac{1}{2}$
C.$\frac{2}{3}$
D.$\frac{1}{6}$

Concept Applied

Prime numbers on a die: 2, 3, 5 → 3 outcomes. P = 3/6 = 1/2....

Read Full Step-by-Step Solution →

Question #10

Practice Question

A.0.2304
B.0.3125
C.0.2560
D.0.2000

Concept Applied

Using the binomial formula P = C(5,3)·(0.4)³·(0.6)² = 10·0.064·0.36 = 0.2304....

Read Full Step-by-Step Solution →

Question #11

Practice Question

Concept Applied

For a binomial distribution, mean $ = np = 8 $, variance $ = np(1-p) = 4 $. Dividing, $ \frac{np}{np(1-p)} = \frac{8}{4} \Rightarrow \frac{1}{1-p} = 2...

Read Full Step-by-Step Solution →

Question #12

Practice Question

A.$\frac{5}{16}$
B.$\frac{3}{8}$
C.$\frac{1}{2}$
D.$\frac{10}{32}$

Concept Applied

Use binomial distribution: $P(X=3) = \binom{5}{3} \left(\frac{1}{2}\right)^5 = \frac{10}{32} = \frac{5}{16}$....

Read Full Step-by-Step Solution →

Question #13

Practice Question

A.The experimental probability must equal the theoretical probability for any number of trials.
B.The experimental probability approaches the theoretical probability as the number of trials increases.
C.The experimental probability is always greater than the theoretical probability.
D.The experimental probability is independent of the number of trials.

Concept Applied

Experimental probability is the ratio of observed favorable outcomes to total trials. As the number of trials grows, the law of large numbers ensures ...

Read Full Step-by-Step Solution →

Question #14

Practice Question

A.Experimental probability equals theoretical probability.
B.Experimental probability is greater than theoretical probability.
C.Experimental probability is less than theoretical probability.
D.Experimental probability cannot be determined from the given data.

Concept Applied

Theoretical probability of drawing a red ball is \(\frac{3}{10}=0.3\). Experimental probability is \(\frac{12}{40}=0.3\). Since both values are equal,...

Read Full Step-by-Step Solution →

Question #15

Practice Question

A.0.224
B.0.149
C.0.332
D.0.500

Concept Applied

For a Poisson distribution with mean \(\lambda = 3\), \(P(X=2)=\frac{e^{-\lambda}\lambda^{2}}{2!}=e^{-3}\frac{9}{2}=4.5e^{-3}\approx0.224\)....

Read Full Step-by-Step Solution →

Question #16

Practice Question

A.7/66
B.13/66
C.19/66
D.1/3

Concept Applied

Total ways to choose 2 balls from 12 is $\binom{12}{2}=66$. Same‑colour selections: red $\binom{5}{2}=10$, blue $\binom{4}{2}=6$, green $\binom{3}{2}=...

Read Full Step-by-Step Solution →

Question #17

Practice Question

A.$\frac{19}{66}$
B.$\frac{1}{3}$
C.$\frac{5}{12}$
D.$\frac{7}{22}$

Concept Applied

Total ways to choose 2 balls: $\binom{12}{2}=66$. Same‑colour ways: $\binom{3}{2}+\binom{4}{2}+\binom{5}{2}=3+6+10=19$. Probability =

9/66$....

Read Full Step-by-Step Solution →

Question #18

Practice Question

Concept Applied

Mean = $np = 4$, Variance = $npq = 3$. Solving gives $q = \frac{3}{4}$, so $p = \frac{1}{4}$, $n = \frac{4}{1/4} = 16$....

Read Full Step-by-Step Solution →

Question #19

Practice Question

A.$\frac{1}{12}$
B.$\frac{1}{6}$
C.$\frac{1}{4}$
D.$\frac{1}{3}$

Concept Applied

The coin can give 0 (tails) or 1 (heads). To get total 7, we need die = 7 (impossible) with heads = 0, or die = 6 with heads = 1. Only one favorable o...

Read Full Step-by-Step Solution →

Question #20

Practice Question

A.$\binom{n}{r} p^r (1-p)^{n-r}$
B.$\binom{n}{r} p^{n-r} (1-p)^r$
C.$\binom{n}{r} p^r (1-p)^r$
D.$\binom{n}{r} p^{n-r} (1-p)^{n-r}$

Concept Applied

Step 1: Use the binomial distribution formula to find the probability that exactly $r$ successes occur in $n$ trials.,Step 2: Simplify the expression ...

Read Full Step-by-Step Solution →

Question #21

Practice Question

A.$\frac{1}{6}$
B.$\frac{1}{3}$
C.$\frac{1}{2}$
D.$\frac{2}{3}$

Concept Applied

Only even prime is 2. One outcome out of six ⇒ $\frac{1}{6}$....

Read Full Step-by-Step Solution →

Question #22

Practice Question

A.0.44
B.0.50
C.0.56
D.0.64

Concept Applied

Tails = 100 - 56 = 44; $P = 44/100 = 0.44$....

Read Full Step-by-Step Solution →

Question #23

Practice Question

A.A fundamental principle of Mathematics.
B.A complex derivation in JEEMains syllabus.
C.An experimental observation.
D.A theoretical assumption.

Concept Applied

This is a placeholder question to ensure comprehensive syllabus coverage. The correct answer highlights the fundamental nature of Probability....

Read Full Step-by-Step Solution →

Question #24

Practice Question

A.$\frac{8}{20}$
B.$\frac{12}{20}$
C.$\frac{7}{20}$
D.$\frac{5}{20}$

Concept Applied

Total number of balls = 5 + 8 + 7 = 20. Number of non-white balls = 5 (red) + 7 (black) = 12. So, P(not white) = $\frac{12}{20}$. This simplifies to $...

Read Full Step-by-Step Solution →

Question #25

Practice Question

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

Concept Applied

After removing one red, 2 red and 2 green remain. P(green) = 2/4 = 1/2....

Read Full Step-by-Step Solution →

Question #26

Practice Question

A.$\frac{15}{64}$
B.$\frac{1}{4}$
C.$\frac{5}{16}$
D.$\frac{3}{8}$

Concept Applied

This is a binomial distribution with $n=6$, $p=0.5$, $k=4$. Use formula: $P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}$. So $P(X=4) = \binom{6}{4} (0.5)^4 (0...

Read Full Step-by-Step Solution →

Question #27

Practice Question

Concept Applied

Use P(A∪B) = P(A) + P(B) - P(A)P(B) for independent events....

Read Full Step-by-Step Solution →

Question #28

Practice Question

A.$\text{Mean: } \frac{5}{3}, \text{ Variance: } \frac{10}{9}$
B.$\text{Mean: } \frac{5}{3}, \text{ Variance: } \frac{20}{9}$
C.$\text{Mean: } \frac{5}{9}, \text{ Variance: } \frac{10}{3}$
D.$\text{Mean: } \frac{5}{9}, \text{ Variance: } \frac{20}{3}$

Concept Applied

Step 1: Recall the formula for the mean of a binomial distribution, which is $np$.,Step 2: Recall the formula for the variance of a binomial distribut...

Read Full Step-by-Step Solution →

Question #29

Practice Question

A.$\frac{1}{6}$
B.$\frac{1}{12}$
C.$\frac{1}{9}$
D.$\frac{5}{36}$

Concept Applied

Total outcomes = 36. Favorable outcomes for sum 7: (1,6),(2,5),(3,4),(4,3),(5,2),(6,1) → 6 cases. So, probability = $\frac{6}{36} = \frac{1}{6}$....

Read Full Step-by-Step Solution →

Question #30

Practice Question

A.\frac{1}{2}
B.\frac{1}{3}
C.\frac{2}{3}
D.\frac{5}{6}

Concept Applied

Even numbers on a die are {2,4,6} → P(A)=3/6=1/2. Complement probability =1−1/2=1/2....

Read Full Step-by-Step Solution →

Question #31

Practice Question

A.1/2
B.1/3
C.2/3
D.5/6

Concept Applied

Even numbers on a die are {2,4,6} (3 outcomes). The complement (odd numbers) also has 3 outcomes. Probability = 3/6 = 1/2....

Read Full Step-by-Step Solution →

Question #32

Practice Question

A.$\frac{1}{3}$
B.$\frac{1}{2}$
C.$\frac{2}{3}$
D.$\frac{1}{6}$

Concept Applied

Prime numbers on a die: 2, 3, 5 → 3 outcomes. P = 3/6 = 1/2....

Read Full Step-by-Step Solution →

Question #33

Practice Question

A.0.2304
B.0.3125
C.0.2560
D.0.2000

Concept Applied

Using the binomial formula P = C(5,3)·(0.4)³·(0.6)² = 10·0.064·0.36 = 0.2304....

Read Full Step-by-Step Solution →

Question #34

Practice Question

Concept Applied

For a binomial distribution, mean $ = np = 8 $, variance $ = np(1-p) = 4 $. Dividing, $ \frac{np}{np(1-p)} = \frac{8}{4} \Rightarrow \frac{1}{1-p} = 2...

Read Full Step-by-Step Solution →

Question #35

Practice Question

A.$\frac{5}{16}$
B.$\frac{3}{8}$
C.$\frac{1}{2}$
D.$\frac{10}{32}$

Concept Applied

Use binomial distribution: $P(X=3) = \binom{5}{3} \left(\frac{1}{2}\right)^5 = \frac{10}{32} = \frac{5}{16}$....

Read Full Step-by-Step Solution →

Question #36

Practice Question

A.The experimental probability must equal the theoretical probability for any number of trials.
B.The experimental probability approaches the theoretical probability as the number of trials increases.
C.The experimental probability is always greater than the theoretical probability.
D.The experimental probability is independent of the number of trials.

Concept Applied

Experimental probability is the ratio of observed favorable outcomes to total trials. As the number of trials grows, the law of large numbers ensures ...

Read Full Step-by-Step Solution →

Question #37

Practice Question

A.Experimental probability equals theoretical probability.
B.Experimental probability is greater than theoretical probability.
C.Experimental probability is less than theoretical probability.
D.Experimental probability cannot be determined from the given data.

Concept Applied

Theoretical probability of drawing a red ball is \(\frac{3}{10}=0.3\). Experimental probability is \(\frac{12}{40}=0.3\). Since both values are equal,...

Read Full Step-by-Step Solution →

Question #38

Practice Question

A.0.224
B.0.149
C.0.332
D.0.500

Concept Applied

For a Poisson distribution with mean \(\lambda = 3\), \(P(X=2)=\frac{e^{-\lambda}\lambda^{2}}{2!}=e^{-3}\frac{9}{2}=4.5e^{-3}\approx0.224\)....

Read Full Step-by-Step Solution →

Question #39

Practice Question

A.7/66
B.13/66
C.19/66
D.1/3

Concept Applied

Total ways to choose 2 balls from 12 is $\binom{12}{2}=66$. Same‑colour selections: red $\binom{5}{2}=10$, blue $\binom{4}{2}=6$, green $\binom{3}{2}=...

Read Full Step-by-Step Solution →

Question #40

Practice Question

A.$\frac{19}{66}$
B.$\frac{1}{3}$
C.$\frac{5}{12}$
D.$\frac{7}{22}$

Concept Applied

Total ways to choose 2 balls: $\binom{12}{2}=66$. Same‑colour ways: $\binom{3}{2}+\binom{4}{2}+\binom{5}{2}=3+6+10=19$. Probability =

9/66$....

Read Full Step-by-Step Solution →

Question #41

Practice Question

Concept Applied

Mean = $np = 4$, Variance = $npq = 3$. Solving gives $q = \frac{3}{4}$, so $p = \frac{1}{4}$, $n = \frac{4}{1/4} = 16$....

Read Full Step-by-Step Solution →

Question #42

Practice Question

A.$\frac{1}{12}$
B.$\frac{1}{6}$
C.$\frac{1}{4}$
D.$\frac{1}{3}$

Concept Applied

The coin can give 0 (tails) or 1 (heads). To get total 7, we need die = 7 (impossible) with heads = 0, or die = 6 with heads = 1. Only one favorable o...

Read Full Step-by-Step Solution →

Question #43

Practice Question

A.$\binom{n}{r} p^r (1-p)^{n-r}$
B.$\binom{n}{r} p^{n-r} (1-p)^r$
C.$\binom{n}{r} p^r (1-p)^r$
D.$\binom{n}{r} p^{n-r} (1-p)^{n-r}$

Concept Applied

Step 1: Use the binomial distribution formula to find the probability that exactly $r$ successes occur in $n$ trials.,Step 2: Simplify the expression ...

Read Full Step-by-Step Solution →

Question #44

Practice Question

A.$\frac{1}{6}$
B.$\frac{1}{3}$
C.$\frac{1}{2}$
D.$\frac{2}{3}$

Concept Applied

Only even prime is 2. One outcome out of six ⇒ $\frac{1}{6}$....

Read Full Step-by-Step Solution →

Question #45

Practice Question

A.0.44
B.0.50
C.0.56
D.0.64

Concept Applied

Tails = 100 - 56 = 44; $P = 44/100 = 0.44$....

Read Full Step-by-Step Solution →

Question #46

Practice Question

A.A fundamental principle of Mathematics.
B.A complex derivation in JEEMains syllabus.
C.An experimental observation.
D.A theoretical assumption.

Concept Applied

This is a placeholder question to ensure comprehensive syllabus coverage. The correct answer highlights the fundamental nature of Probability....

Read Full Step-by-Step Solution →

Question #47

Practice Question

A.$\frac{8}{20}$
B.$\frac{12}{20}$
C.$\frac{7}{20}$
D.$\frac{5}{20}$

Concept Applied

Total number of balls = 5 + 8 + 7 = 20. Number of non-white balls = 5 (red) + 7 (black) = 12. So, P(not white) = $\frac{12}{20}$. This simplifies to $...

Read Full Step-by-Step Solution →

Question #48

Practice Question

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

Concept Applied

After removing one red, 2 red and 2 green remain. P(green) = 2/4 = 1/2....

Read Full Step-by-Step Solution →

Question #49

Practice Question

A.$\frac{15}{64}$
B.$\frac{1}{4}$
C.$\frac{5}{16}$
D.$\frac{3}{8}$

Concept Applied

This is a binomial distribution with $n=6$, $p=0.5$, $k=4$. Use formula: $P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}$. So $P(X=4) = \binom{6}{4} (0.5)^4 (0...

Read Full Step-by-Step Solution →

Question #50

Practice Question

Concept Applied

Use P(A∪B) = P(A) + P(B) - P(A)P(B) for independent events....

Read Full Step-by-Step Solution →
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