JEE Main Previous Year Probability Questions and Solutions | Addition Theorem, Conditional Probability and Bayes’ Theorem

Q 11 :
An integer is chosen at random from the integers 1, 2, 3, ..., 50. The probability that the chosen integer is a multiple of at least one of 4, 6, and 7 is [2024]

$\frac{8}{25}$
$\frac{9}{50}$
$\frac{14}{25}$
$\frac{21}{50}$

1

2

3

4

(4)

Let A, B, C be the events represents the numbers divisible by 4, 6 and 7 respectively.

$∴$ $n (A) = 12, n (B) = 8, n (C) = 7$

$n (A \cap B) = 4, n (B \cap C) = 1, n (A \cap C) = 1$

$and n (A \cap B \cap C) = 0$

$So, n (A \cup B \cup C) = 12 + 8 + 7 - 4 - 1 - 1 = 21$

$∴$ $Required probability = \frac{21}{50}$

Q 12 :
Two integers $x$ and $y$ are chosen with replacement from the set {0, 1, 2, 3,…,10}. Then the probability that $| x - y | > 5,$ is [2024]

$\frac{60}{121}$
$\frac{31}{121}$
$\frac{62}{121}$
$\frac{30}{121}$

1

2

3

4

(4)

Let $E = {(x, y) : | x - y | > 5}$ and $x, y \in {0, 1, 2, \dots, 10}$

$∴$ $E = {(0, 6), (0, 7), (0, 8), (0, 9), (0, 10), (1, 7), (1, 8), (1, 9), (1, 10), (2, 8), (2, 9), (2, 10), (3, 9), (3, 10), (4, 10),$ $(6, 0), (7, 0), (8, 0), (9, 0), (10, 0), (7, 1), (8, 1), (9, 1), (10, 1), (8, 2), (9, 2), (10, 2), (9, 3), (10, 3), (10, 4)}$

$∴$ $n (E) = 30$

Total number of possible outcomes $= 11 \times 11 = 121$

$∴$ $Required probability = \frac{30}{121}$

Q 13 :
Bag A contains 3 white, 7 red balls and bag B contains 3 white, 2 red balls. One bag is selected at random and a ball is drawn from it. The probability of drawing the ball from bag A, if the ball drawn is white, is [2024]

$\frac{3}{10}$
$\frac{1}{4}$
$\frac{1}{9}$
$\frac{1}{3}$

1

2

3

4

(4)

Let $E_{1}$ be the event that bag A is selected, $E_{2}$ be the event that bag B is selected and E be the event that white ball is drawn.

$∴$ $P (E_{1}) = P (E_{2}) = \frac{1}{2}, P (\frac{E}{E_{1}}) = \frac{3}{10}, P (\frac{E}{E_{2}}) = \frac{3}{5}$

$∴ Required probability,$

$P (\frac{E_{1}}{E}) = \frac{P (E_{1}) \cdot P (\frac{E}{E_{1}})}{P (E_{1}) \cdot P (\frac{E}{E_{1}}) + P (E_{2}) \cdot P (\frac{E}{E_{2}})}$

$= \frac{\frac{1}{2} \times \frac{3}{10}}{\frac{1}{2} \times \frac{3}{10} + \frac{1}{2} \times \frac{3}{5}} = \frac{1}{3}$

Q 14 :
Two marbles are drawn in succession from a box containing 10 red, 30 white, 20 blue and 15 orange marbles, with replacement being made after each drawing. Then the probability that the first drawn marble is red and the second drawn marble is white, is [2024]

$\frac{2}{25}$
$\frac{4}{75}$
$\frac{2}{3}$
$\frac{4}{25}$

1

2

3

4

(2)

Total marbles = 10 + 30 + 20 + 15 = 75

Let $E$ be the event of drawing first drawn marble is red and second drawn marble is white.

Now, probability of drawing first red marble and $2^{n d}$ white marble is

$P (E) = \frac{10}{75} \times \frac{30}{75} = \frac{4}{75}$

Q 15 :
A coin is biased so that a head is twice as likely to occur as a tail. If the coin is tossed 3 times, then the probability of getting two tails and one head is [2024]

$\frac{2}{27}$
$\frac{2}{9}$
$\frac{1}{27}$
$\frac{1}{9}$

1

2

3

4

(2)

Let $x$ be the probability of getting a tail.

$∴$ $P (T) = x; P (H) = 2 x$

$∵$ $x + 2 x = 1 \Rightarrow x = \frac{1}{3}$

$∴$ $P (T) = \frac{1}{3}$ and $P (H) = \frac{2}{3}$

$Now, required probability = {}^{3}{C_{2}} \times \frac{1}{3} \times \frac{1}{3} \times \frac{2}{3}$

$= 3 \times \frac{2}{27} = \frac{2}{9}$

Q 16 :
In a tournament, a team plays 10 matches with probabilities of winning and losing each match as $\frac{1}{3}$ and $\frac{2}{3}$ respectively. Let $x$ be the number of matches that the team wins, and $y$ be the number of matches that the team loses. If the probability $P (∣ x - y ∣ \leq 2)$ is $p$ , then $3^{9} p$ equals _______ . [2024]

0%

(8288)

$| x - y | \leq 10$ and $0 \leq x \leq 10$ and $0 \leq y \leq 10$

$∴$ $p = P (| x - y | \leq 2)$

$= P (x = 4, y = 6) + P (x = 5, y = 5) + P (x = 6, y = 4)$

$= {}^{10}{C_{4}} {(\frac{1}{3})}^{4} {(\frac{2}{3})}^{6} + {}^{10}{C_{5}} {(\frac{1}{3})}^{5} {(\frac{2}{3})}^{5} + {}^{10}{C_{6}} {(\frac{1}{3})}^{6} {(\frac{2}{3})}^{4}$

$= 210 (\frac{2^{6}}{3^{10}}) + 252 (\frac{2^{5}}{3^{10}}) + 210 (\frac{2^{4}}{3^{10}})$

$= \frac{2^{5}}{3^{10}} [420 + 252 + 105] \Rightarrow p = \frac{2^{5}}{3^{10}} \times 777$

$Hence, 3^{9} p = 3^{9} \times \frac{2^{5}}{3^{10}} \times 777 = \frac{2^{5}}{3} \times 777 = 8288$

Q 17 :
Let the mean and the standard deviation of the probability distribution

X $α$ 1 0 $- 3$

P(X) $\frac{1}{3}$ $K$ $\frac{1}{6}$ $\frac{1}{4}$

be $μ$ and $σ$ , respectively. If $σ - μ = 2,$ then $σ + μ$ is equal to _____________. [2024]

0%

(5)

We have, $\sum P (X) = 1$

$\Rightarrow \frac{1}{3} + K + \frac{1}{6} + \frac{1}{4} = 1 \Rightarrow \frac{8 + 4 + 6}{24} + K = 1$

$\Rightarrow K = 1 - \frac{18}{24} = \frac{6}{24} = \frac{1}{4}$

Now, $μ = \frac{α}{3} + \frac{1}{4} - \frac{3}{4} = \frac{α}{3} - \frac{1}{2}$

and $σ = \sqrt{(\frac{α^{2}}{3} + \frac{1}{4} + \frac{9}{4}) - {(\frac{α}{3} - \frac{1}{2})}^{2}}$ $= \sqrt{(\frac{2 α^{2}}{9} + \frac{α}{3} + \frac{9}{4})}$

Given, $σ - μ = 2$

$\Rightarrow σ = μ + 2$

$\Rightarrow σ^{2} = {(μ + 2)}^{2}$ $\Rightarrow \frac{2 α^{2}}{9} + \frac{α}{3} + \frac{9}{4} = \frac{α^{2}}{9} + \frac{9}{4} + α$

$\Rightarrow \frac{α^{2}}{9} - \frac{2 α}{3} = 0$

$\Rightarrow α = 0, 6$ $[α \neq 0$ $∵$ $X = 0$ is already given $]$

$\Rightarrow α = 6 \Rightarrow μ = \frac{3}{2}$ and $σ = \frac{7}{2}$

$So μ + σ = 5$

Q 18 :
Let $a, b$ and $c$ denote the outcome of three independent rolls of a fair tetrahedral die, whose four faces are marked 1, 2, 3, 4. If the probability that $a x^{2} + b x + c = 0$ has all real roots is $\frac{m}{n}, g c d (m, n) = 1,$ then $m + n$ is equal to _______ . [2024]

0%

(19)

We have, $a x^{2} + b x + c = 0$

For real roots, $b^{2} - 4 a c \geq 0$ ...(i)

$a, b, c \in {1, 2, 3, 4}$ ...(ii)

Ordered triplet $(a, b, c)$ satisfying (i) and (ii) are

$(1, 2, 1), (1, 3, 1), (2, 3, 1), (1, 3, 2), (1, 4, 1), (1, 4, 2), (2, 4, 1), (2, 4, 2), (4, 4, 1), (1, 4, 4), (3, 4, 1), (1, 4, 3)$

i.e. total 12 favourable outcomes.

Total number of outcomes = $4 \times 4 \times 4 = 64$

$∴$ $Required probability = \frac{12}{64} = \frac{3}{16} = \frac{m}{n}$ $(∵ Given)$

Here, $m + n = 3 + 16 = 19$

Q 19 :
Given three identical balls each containing 10 balls, whose colours are as follows:

Red Blue Green

Bag I 3 2 5

Bag II 4 3 3

Bag III 5 1 4

A person chooses a bag at random and takes out a ball. If the ball is Red, the probability that it is from bag I is p and if the ball is Green, the probability that it is from bag III is q, then the value of $(\frac{1}{p} + \frac{1}{q})$ is : [2025]

6
9
7
8

1

2

3

4

(3)

Let us define the Events

A : Ball drawn from bag I

B : Ball drawn from bag II

C : Ball drawn from bag III

R : Red ball is drawn

G : Green ball is drawn

$∴ P (R) = P (A) \cdot P (R / A) + P (B) \cdot P (R / B) + P (C) \cdot P (R / C)$

$= \frac{1}{3} \times \frac{3}{10} + \frac{1}{3} \times \frac{4}{10} + \frac{1}{3} \times \frac{5}{10} = \frac{12}{30}$

Now, $p = P (A / R) = \frac{P (A) P (R / A)}{P (R)} = \frac{\frac{1}{3} \times \frac{3}{10}}{\frac{12}{30}} = \frac{3}{12} = \frac{1}{4}$

Now, $P (G) = \frac{1}{3} \times \frac{5}{10} + \frac{1}{3} \times \frac{3}{10} + \frac{1}{3} \times \frac{4}{10} = \frac{12}{30}$

$∴ q = P (C / G) = \frac{P (C) \cdot P (G / C)}{P (G)} = \frac{\frac{1}{3} \times \frac{4}{10}}{\frac{12}{30}} = \frac{4}{12} = \frac{1}{3}$

$∴ \frac{1}{p} + \frac{1}{q} = 4 + 3 = 7$ .

Q 20 :
The probability, of forming a 12 persons committee from 4 engineers, 2 doctors and 10 professors containing at least 3 engineers and at least 1 doctor, is [2025]

$\frac{129}{182}$
$\frac{103}{182}$
$\frac{17}{26}$
$\frac{19}{26}$

1

2

3

4

(1)

Total number of ways of selecting 12 people from 4 engineers, 2 doctors and 10 professors = ${}^{16}C_{12}$ = 1820.

To determine favourite outcomes:-

To select at least 3 engineers 1 doctor is, we can choose

4 Engineers	2 Doctors	10 Professors	Number of ways
3	1	8	${}^{4}C_{3} \times {}^{2}C_{1} \times {}^{10}C_{8} = 360$
4	2	6	${}^{4}C_{4} \times {}^{2}C_{2} \times {}^{10}C_{6} = 210$
3	2	7	${}^{4}C_{3} \times {}^{2}C_{2} \times {}^{10}C_{7} = 480$
4	1	7	${}^{4}C_{4} \times {}^{2}C_{1} \times {}^{10}C_{7} = 240$

Total favourable outcomes = 1290

So, required probability $= \frac{1290}{1820} = \frac{129}{182}$ .

Topic Question Set

Q 11 :
An integer is chosen at random from the integers 1, 2, 3, ..., 50. The probability that the chosen integer is a multiple of at least one of 4, 6, and 7 is [2024]

Q 12 :
Two integers $x$ and $y$ are chosen with replacement from the set {0, 1, 2, 3,…,10}. Then the probability that $| x - y | > 5,$ is [2024]

Q 13 :
Bag A contains 3 white, 7 red balls and bag B contains 3 white, 2 red balls. One bag is selected at random and a ball is drawn from it. The probability of drawing the ball from bag A, if the ball drawn is white, is [2024]

Q 14 :
Two marbles are drawn in succession from a box containing 10 red, 30 white, 20 blue and 15 orange marbles, with replacement being made after each drawing. Then the probability that the first drawn marble is red and the second drawn marble is white, is [2024]

Q 15 :
A coin is biased so that a head is twice as likely to occur as a tail. If the coin is tossed 3 times, then the probability of getting two tails and one head is [2024]

0%

Q 17 :
Let the mean and the standard deviation of the probability distribution

X $α$ 1 0 $- 3$

P(X) $\frac{1}{3}$ $K$ $\frac{1}{6}$ $\frac{1}{4}$

be $μ$ and $σ$ , respectively. If $σ - μ = 2,$ then $σ + μ$ is equal to _____________. [2024]

0%

Q 18 :
Let $a, b$ and $c$ denote the outcome of three independent rolls of a fair tetrahedral die, whose four faces are marked 1, 2, 3, 4. If the probability that $a x^{2} + b x + c = 0$ has all real roots is $\frac{m}{n}, g c d (m, n) = 1,$ then $m + n$ is equal to _______ . [2024]

0%

Q 20 :
The probability, of forming a 12 persons committee from 4 engineers, 2 doctors and 10 professors containing at least 3 engineers and at least 1 doctor, is [2025]

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X	$α$	1	0	$- 3$
P(X)	$\frac{1}{3}$	$K$	$\frac{1}{6}$	$\frac{1}{4}$

Topic Question Set

Q 11 : An integer is chosen at random from the integers 1, 2, 3, ..., 50. The probability that the chosen integer is a multiple of at least one of 4, 6, and 7 is [2024]

Q 12 : Two integers x and y are chosen with replacement from the set {0, 1, 2, 3,…,10}. Then the probability that |x-y|>5, is [2024]

Q 13 : Bag A contains 3 white, 7 red balls and bag B contains 3 white, 2 red balls. One bag is selected at random and a ball is drawn from it. The probability of drawing the ball from bag A, if the ball drawn is white, is [2024]

Q 14 : Two marbles are drawn in succession from a box containing 10 red, 30 white, 20 blue and 15 orange marbles, with replacement being made after each drawing. Then the probability that the first drawn marble is red and the second drawn marble is white, is [2024]

Q 15 : A coin is biased so that a head is twice as likely to occur as a tail. If the coin is tossed 3 times, then the probability of getting two tails and one head is [2024]

0%

Q 17 : Let the mean and the standard deviation of the probability distribution X α 1 0 -3 P(X) 13 K 16 14 be μ and σ, respectively. If σ-μ=2, then σ+μ is equal to _____________. [2024]

0%

Q 18 : Let a,b and c denote the outcome of three independent rolls of a fair tetrahedral die, whose four faces are marked 1, 2, 3, 4. If the probability that ax2+bx+c=0 has all real roots is mn,gcd(m,n)=1, then m+n is equal to _______ . [2024]

0%

Q 20 : The probability, of forming a 12 persons committee from 4 engineers, 2 doctors and 10 professors containing at least 3 engineers and at least 1 doctor, is [2025]

Want Us to Email you About Special Offers & Updates?

Q 11 :
An integer is chosen at random from the integers 1, 2, 3, ..., 50. The probability that the chosen integer is a multiple of at least one of 4, 6, and 7 is [2024]

Q 12 :
Two integers $x$ and $y$ are chosen with replacement from the set {0, 1, 2, 3,…,10}. Then the probability that $| x - y | > 5,$ is [2024]

Q 13 :
Bag A contains 3 white, 7 red balls and bag B contains 3 white, 2 red balls. One bag is selected at random and a ball is drawn from it. The probability of drawing the ball from bag A, if the ball drawn is white, is [2024]

Q 14 :
Two marbles are drawn in succession from a box containing 10 red, 30 white, 20 blue and 15 orange marbles, with replacement being made after each drawing. Then the probability that the first drawn marble is red and the second drawn marble is white, is [2024]

Q 15 :
A coin is biased so that a head is twice as likely to occur as a tail. If the coin is tossed 3 times, then the probability of getting two tails and one head is [2024]

Q 17 :
Let the mean and the standard deviation of the probability distribution

X $α$ 1 0 $- 3$

P(X) $\frac{1}{3}$ $K$ $\frac{1}{6}$ $\frac{1}{4}$

be $μ$ and $σ$ , respectively. If $σ - μ = 2,$ then $σ + μ$ is equal to _____________. [2024]

Q 18 :
Let $a, b$ and $c$ denote the outcome of three independent rolls of a fair tetrahedral die, whose four faces are marked 1, 2, 3, 4. If the probability that $a x^{2} + b x + c = 0$ has all real roots is $\frac{m}{n}, g c d (m, n) = 1,$ then $m + n$ is equal to _______ . [2024]

Q 20 :
The probability, of forming a 12 persons committee from 4 engineers, 2 doctors and 10 professors containing at least 3 engineers and at least 1 doctor, is [2025]