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

Q 1 :

Three urns A, B and C contain 7 red, 5 black, 5 red, 7 black and 6 red, 6 black balls, respectively. One of the urns is selected at random and a ball is drawn from it. If the ball drawn is black, then the probability that it is drawn from urn A is __________ . [2024]

$\frac{5}{16}$
$\frac{4}{17}$
$\frac{5}{18}$
$\frac{7}{18}$

1

2

3

4

(3)

Let A, B, C and E be the events defined as follows

A: First urn is chosen

B: Second urn is chosen

C: Third urn is chosen

E: Ball drawn is black

$∴$ $P (A) = P (B) = P (C) = \frac{1}{3}$

$∴$ $P (E | A) = \frac{5}{12}, P (E | B) = \frac{7}{12}, P (E | C) = \frac{6}{12}$

By Bayes' Theorem,

$P (A | E) = \frac{P (A) \cdot P (E | A)}{P (A) \cdot P (E | A) + P (B) \cdot P (E | B) + P (C) \cdot P (E | C)}$

$= \frac{\frac{1}{3} \times \frac{5}{12}}{\frac{1}{3} \times \frac{5}{12} + \frac{1}{3} \times \frac{7}{12} + \frac{1}{3} \times \frac{6}{12}} = \frac{5}{5 + 7 + 6} = \frac{5}{18}$

Q 2 :

A company has two plants A and B to manufacture motorcycles. 60% motorcycles are manufactured at plant A and the remaining are manufactured at plant B. 80% of the motorcycles manufactured at plant A are rated of the standard quality, while 90% of the motorcycles manufactured at plant B are rated of the standard quality. A motorcycle picked up randomly from the total production is found to be of the standard quality. If $p$ is the probability that it was manufactured at plant B, then $126 p$ is [2024]

64
56
66
54

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4

(4)

Let $E_{1}$ be the event that motorcycle manufactured in plant A.

$E_{2}$ be the event that motorcycle manufactured in plant B, and E be the event that it found to be of standard quality.

$∴$ $P (E_{1}) = \frac{60}{100}, P (E_{2}) = \frac{40}{100}$

$P (\frac{E}{E_{1}}) = \frac{80}{100}$ and $P (\frac{E}{E_{2}}) = \frac{90}{100}$

Now, $p = P (\frac{E_{2}}{E}) = \frac{P (E | E_{2}) \cdot P (E_{2})}{P (E | E_{2}) \cdot P (E_{2}) + P (E | E_{1}) \cdot (E_{1})}$

$= \frac{\frac{90}{100} \cdot \frac{40}{100}}{\frac{60}{100} \cdot \frac{80}{100} + \frac{90}{100} \cdot \frac{40}{100}} = \frac{36}{48 + 36} = \frac{3}{7}$

$∴$ $126 p = 126 \times \frac{3}{7} = 54$

Q 3 :

If three letters can be posted to any one of the 5 different addresses, then the probability that the three letters are posted to exactly two addresses is ______. [2024]

$\frac{6}{25}$
$\frac{4}{25}$
$\frac{18}{25}$
$\frac{12}{25}$

1

2

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4

(4)

We need to choose 2 addresses out of 5 which can be done in ${}^{5}C_{2}$ ways.

Now, 3 letters can be posted to exactly 2 addresses in $3 \times 2 = 6$ ways.

Required Probability = $\frac{{}^{5}{C_{2}} \times 6}{5^{3}} = \frac{60}{125} = \frac{12}{25}$

Q 4 :

Let the sum of two positive integers be 24. If the probability that their product is not less than $\frac{3}{4}$ times their greatest possible product is $\frac{m}{n},$ where $g c d (m, n) = 1,$ then $n - m$ equals [2024]

9
10
11
8

1

2

3

4

(2)

Let the two positive integers be $x$ and $y$ .

Now, $x + y = 24,$ $x, y \in N$

Now, $\frac{x + y}{2} \geq \sqrt{x y}$ $[∵ A . M . \geq G . M .]$

$\Rightarrow x y \leq 144$

Since, $\frac{3}{4} \times 144 = 108$

So, $x y \geq 108$

$∴$ Favourable cases = ${(6, 18), (18, 6), (7, 17), (17, 7), (8, 16), (16, 8), (9, 15), (15, 9), (10, 14), (14, 10), (11, 13), (13, 11), (12, 12)}$

Total choices for $x + y = 24$ are 23

$∴$ Required probability $= \frac{13}{23} = \frac{m}{n}$

$∴$ $n - m = 23 - 13 = 10$

Q 5 :

There are three bags X, Y, and Z. Bag X contains 5 one-rupee coins and 4 five-rupee coins; Bag Y contains 4 one-rupee coins and 5 five-rupee coins, and Bag Z contains 3 one-rupee coins and 6 five-rupee coins. A bag is selected at random and a coin drawn from it at random is found to be a one-rupee coin. Then the probability that it came from bag Y is [2024]

$\frac{1}{2}$
$\frac{5}{12}$
$\frac{1}{4}$
$\frac{1}{3}$

1

2

3

4

(4)

Let $E_{1}, E_{2}, E_{3}$ be the event of selecting bags X, Y and Z respectively and A be the event that coin drawn is one-rupee coin.

$∴$ $P (E_{1}) = P (E_{2}) = P (E_{3}) = \frac{1}{3}$

$P (A | E_{1}) = \frac{5}{9}, P (A | E_{2}) = \frac{4}{9}, P (A | E_{3}) = \frac{3}{9}$

By Bayes' theorem,

Required Probability = $P (E_{2} | A) = \frac{\frac{1}{3} \times \frac{4}{9}}{\frac{1}{3} \times \frac{5}{9} + \frac{1}{3} \times \frac{4}{9} + \frac{1}{3} \times \frac{3}{9}}$

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

Q 6 :

If an unbiased dice is rolled thrice, then the probability of getting a greater number in the $i^{t h}$ roll than the number obtained in the ${(i - 1)}^{t h}$ roll, $i = 2, 3,$ is equal to [2024]

$\frac{5}{54}$
$\frac{2}{54}$
$\frac{1}{54}$
$\frac{3}{54}$

1

2

3

4

(1)

Let X be the event of getting a greater number than the previous one in a throw of a die.

Favourable outcomes to $X = {}^{6}{C_{3}}$

[ $∵$ ${}^{6}{C_{3}}$ are ways of getting 3 outcomes one less than other as we have total 6 possible outcomes]

Total outcomes = $6^{3}$

$P (X) = \frac{{}^{6}{C_{3}}}{6^{3}} = \frac{20}{216} = \frac{5}{54}$

Q 7 :

A bag contains 8 balls, whose colours are either white or black. 4 balls are drawn at random without replacement and it was found that 2 balls are white and other 2 balls are black. The probability that the bag contains equal number of white and black balls is [2024]

$\frac{2}{5}$
$\frac{1}{7}$
$\frac{1}{5}$
$\frac{2}{7}$

1

2

3

4

(4)

The balls can be distributed as (2W, 6B), (3W, 5B), (4W, 4B), (5W, 3B), (6W, 2B)

$∴$ $Required probability$ $= \frac{\frac{{}^{4}C_{2} \times {}^{4}{C_{2}}}{\frac{{}^{8}{C_{4}}}{2 ({}^{2}{C_{2} \times {}^{6}{C_{2}}}) + (2) ({}^{3}{C_{2} \times {}^{5}{C_{2}}}) + {}^{4}{C_{2}} \times {}^{4}{C_{2}}}}}{{}^{8}{C_{4}}}$

$= \frac{6 \times 6}{2 (15) + 2 (30) + 36} = \frac{36}{30 + 60 + 36} = \frac{36}{126} = \frac{2}{7}$

Q 8 :

Let Ajay will not appear in JEE exam with probability $p = \frac{2}{7},$ while both Ajay and Vijay will appear in the exam with probability $q = \frac{1}{5} .$ Then the probability that Ajay will appear in the exam and Vijay will not appear is [2024]

$\frac{9}{35}$
$\frac{3}{35}$
$\frac{24}{35}$
$\frac{18}{35}$

1

2

3

4

(4)

We have, $P (Not Ajay) = P = \frac{2}{7} \Rightarrow P (Ajay) = P = \frac{5}{7}$

$P (Ajay \cap Vijay) = q = \frac{1}{5}$

$∴$ $P (Ajay \cap not Vijay) = \bar{p} - q = \frac{5}{7} - \frac{1}{5} = \frac{18}{35}$

Q 9 :

An urn contains 6 white and 9 black balls. Two successive draws of 4 balls are made without replacement. The probability that the first draw gives all white balls and the second draw gives all black balls, is [2024]

$\frac{5}{715}$
$\frac{5}{256}$
$\frac{3}{715}$
$\frac{3}{256}$

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2

3

4

(3)

Let A : event when $1^{s t}$ draw gives all 4 white balls.

and B : event when $2^{n d}$ draw gives all 4 black balls.

$P (A) = \frac{{}^{6}{C_{4}}}{{}^{15}{C_{4}}} = \frac{1}{91}$ and $P (B | A) = \frac{{}^{9}{C_{4}}}{{}^{11}{C_{4}}} = \frac{21}{55}$

$∴$ $Required probability = P (A) \times P (\frac{B}{A}) = \frac{1}{91} \times \frac{21}{55} = \frac{3}{715}$

Q 10 :

A fair die is thrown until 2 appears. Then the probability that 2 appears in an even number of throws, is [2024]

$\frac{1}{6}$
$\frac{6}{11}$
$\frac{5}{6}$
$\frac{5}{11}$

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2

3

4

(4)

Probability that 2 appears $= \frac{1}{6}$

Probability that 2 does not appear $= \frac{5}{6}$

Required probability $= \frac{5}{6} \times \frac{1}{6} + \frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} \times \frac{1}{6} + \frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} \times \frac{1}{6} + \dots$

$= \frac{5}{6} \times \frac{1}{6} + {(\frac{5}{6})}^{3} \times \frac{1}{6} + {(\frac{5}{6})}^{5} \times \frac{1}{6} + \dots$ $= \frac{\frac{5}{6} \times \frac{1}{6}}{1 - \frac{5}{6} \times \frac{5}{6}}$ $= \frac{5}{11}$

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]

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

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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]

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

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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]

(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]

(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]

(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 $= \frac{1290}{1820} = \frac{129}{182}$ .

Q 21 :

A bag contains 19 unbiased coins and one coin with head on both sides. One coin drawn at random is tossed and head turns up. If the probability that the drawn coin was unbiased, is $\frac{m}{n}$ , gcd (m, n) = 1, then $n^{2} - m^{2}$ is equal to [2025]

64
80
72
60

1

2

3

4

(2)

Let us defined the events as

A : Unbiased coin is selected

B : Biased coin in selected

H : Head turns up.

$∴ P (H) = P (A) P (H / A) + P (B) P (H / B)$

$= \frac{19}{20} \times \frac{1}{2} + \frac{1}{20} \times 1 = \frac{21}{40}$

Now, $P (A / H) = \frac{P (A) P (H / A)}{P (H)} = \frac{\frac{19}{20} \times \frac{1}{2}}{\frac{21}{40}} = \frac{19}{21}$

$\Rightarrow \frac{m}{n} = \frac{19}{21}$

$∴ n^{2} - m^{2} = 441 - 361 = 80$ .

Q 22 :

If A and B are two events such that P(A) = 0.7, P(B) = 0.4 and $P (A \cap \bar{B}) = 0.5$ , where $\bar{B}$ denotes the complement of B, then $P (B / (A \cup \bar{B}))$ is equal to [2025]

$\frac{1}{3}$
$\frac{1}{2}$
$\frac{1}{6}$
$\frac{1}{4}$

1

2

3

4

(4)

$P (A \cap \bar{B}) = P (A) - P (A \cap B)$

$\Rightarrow P (A \cap B) = P (A) - P (A \cap \bar{B})$

=0.7 – 0.5 = 0.2

Now, $P (A \cup \bar{B}) = P (A) - P (\bar{B}) - P (A \cap \bar{B})$

= 0.7+(1– 0.4) – 0.5 = 0.8

$P (B \cap (A \cup \bar{B})) = P (B \cap A) \cup (B \cap \bar{B})$

$= P (B \cap A) [P (B \cap \bar{B}) = 0]$

= 0.2

Now, $P (B / (A \cup \bar{B})) = \frac{P (B \cap (A \cup \bar{B}))}{P (A \cup \bar{B})} = \frac{0.2}{0.8} = \frac{1}{4}$ .

Q 23 :

Two balls are selected at random one by one without replacement from a bag containing 4 white and 6 black balls. If the probability that the first selected balls is black, given that the second selected ball is also black is $\frac{m}{n}$ , where gcd (m, n) = 1, then m + n is equal to : [2025]

11
13
4
14

1

2

3

4

(4)

Required probability

$= \frac{\frac{6}{10} \times \frac{5}{9}}{\frac{4}{10} \times \frac{6}{9} + \frac{6}{10} \times \frac{5}{9}} = \frac{30}{24 + 30} = \frac{30}{54} = \frac{5}{9}$

So, m = 5, n = 9 [ $∵$ gcd (m, n) = 1]

$∴$ m + n = 14.

Q 24 :

If A and B are two events such that $P (A \cap B) = 0.1$ , and $P (A | B)$ and $P (B | A)$ are the roots of the equation $12 x^{2} - 7 x + 1 = 0$ , then the value of $\frac{P (\bar{A} \cup \bar{B})}{P (\bar{A} \cap \bar{B})}$ is : [2025]

$\frac{5}{3}$
$\frac{4}{3}$
$\frac{7}{4}$
$\frac{9}{4}$

1

2

3

4

(4)

We have, $12 x^{2} - 7 x + 1 = 0$

$\Rightarrow x = \frac{1}{3}, \frac{1}{4}$

Let $P (\frac{A}{B}) = \frac{1}{3} a n d P (\frac{B}{A}) = \frac{1}{4}$

$\Rightarrow \frac{P (A \cap B)}{P (B)} = \frac{1}{3} and \frac{P (A \cap B)}{P (A)} = \frac{1}{4}$

$\Rightarrow P (B) = 0.3 and P (A) = 0.4$

$∴ P (A \cup B) = 0.3 + 0.4 - 0.1 = 0.6$

Now, $\frac{P (\bar{A} \cup \bar{B})}{P (\bar{A} \cap \bar{B})} = \frac{P (\bar{A \cap B})}{P (\bar{A \cup B})}$

$= \frac{1 - P (A \cap B)}{1 - P (A \cup B)} = \frac{1 - 0.1}{1 - 0.6} = \frac{9}{4}$ .

Q 25 :

One die has two faces marked 1, two faces marked 2, one face marked 3 and one face marked 4. Another die has one face marked 1, two faces marked 2, two faces marked 3 and one face marked 4. The probability of getting the sum of numbers to be 4 or 5, when both the dice are thrown together, is [2025]

$\frac{1}{2}$
$\frac{2}{3}$
$\frac{4}{9}$
$\frac{3}{5}$

1

2

3

4

(1)

Let x be the number shows on dice –1 and y be the number shows on dice –2.

Favorable outcomes = {(x, y) = (1, 3), (3, 1), (2, 2), (2, 3), (3, 2), (1, 4), (4, 1)}

$∴$ Required probability

$= \frac{2}{6} \times \frac{2}{6} + \frac{1}{6} \times \frac{1}{6} + \frac{2}{6} \times \frac{2}{6} + \frac{2}{6} \times \frac{2}{6} + \frac{1}{6} \times \frac{2}{6} + \frac{2}{6} \times \frac{1}{6} + \frac{1}{6} \times \frac{1}{6}$

$= \frac{18}{36} = \frac{1}{2}$ .

Q 26 :

A board has 16 squares as shown in the figure:

Out of these 16 squares, two squares are chosen at random. The probability that they have no side in common is: [2025]

$\frac{4}{5}$
$\frac{3}{5}$
$\frac{7}{10}$
$\frac{23}{30}$

1

2

3

4

(1)

Total ways for selecting any two squares $= {}^{16}C_{2} = 120$

Total ways for selecting common side squares $= \underset{Horizontal side}{\underset{⏟}{3 \times 4}} + \underset{Vertical side}{\underset{⏟}{3 \times 4}} = 24$

So, required probability $= 1 - \frac{24}{120} = \frac{96}{120} = \frac{4}{5}$ .

Q 27 :

A and B alternately throw a pair of dice. A wins if he throws a sum of 5 before B throws a sum of 8, and B wins if he throws a sum of 8 before A throws a sum of 5. The probability, that A wins if A makes the first throw, is [2025]

$\frac{8}{17}$
$\frac{9}{19}$
$\frac{9}{17}$
$\frac{8}{19}$

1

2

3

4

(2)

Let $E_{1}$ be the event that A get sum of 5 and $E_{2}$ be the event that B get sum of 8.

$P (E_{1}) = \frac{4}{36} = \frac{1}{9} and P (E_{2}) = \frac{5}{36}$

$∴$ Required Probability

$= P (E_{1}) + P ({\bar{E}}_{1}) P ({\bar{E}}_{2}) P (E_{1}) + P ({\bar{E}}_{1}) P ({\bar{E}}_{2}) P ({\bar{E}}_{1}) P ({\bar{E}}_{2}) P (E_{1}) + . . . \infty$

$= \frac{1}{9} + \frac{8}{9} \times \frac{31}{36} \times \frac{1}{9} + \frac{8}{9} \times \frac{31}{36} \times \frac{8}{9} \times \frac{31}{36} \times \frac{1}{9} + . . . \infty$

$= \frac{\frac{1}{9}}{1 - \frac{62}{81}} = \frac{9}{19}$ .

Q 28 :

Let $A = [a_{i j}]$ be a square matrix of order 2 with entries either 0 or 1. Let E be the event that A is an invertible matrix. Then the probability P(E) is: [2025]

$\frac{5}{8}$
$\frac{3}{8}$
$\frac{3}{16}$
$\frac{1}{8}$

1

2

3

4

(2)

Given : $A = {[a_{i j}]}_{2 \times 2}$ with entries 0 or 1.

Total possible ways for matrix A are $4 \times 4 = 16$ ways.

Here E = { $A | A$ is invertible}

$[\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix}], [\begin{matrix} 1 & 1 \\ 1 & 0 \end{matrix}], [\begin{matrix} 1 & 0 \\ 1 & 1 \end{matrix}], [\begin{matrix} 0 & 1 \\ 1 & 1 \end{matrix}], [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] and [\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}]$ = 6 matrices

Thus, probability P(E) is given by $= \frac{6}{16} = \frac{3}{8}$ .

Q 29 :

Two number $k_{1}$ and $k_{2}$ are randomly chosen from the set of natural numbers. Then, the probability that the value of $i^{k_{1}} + i^{k_{2}}, (i = \sqrt{- 1})$ is non-zero, equals [2025]

$\frac{2}{3}$
$\frac{1}{2}$
$\frac{1}{4}$
$\frac{3}{4}$

1

2

3

4

(4)

Given, $i^{k_{1}} + i^{k_{2}} \neq 0$

Possible values of $i^{k_{1}}$ = 4 = Possible values of $i^{k_{2}}$

$∴$ Cases are i, –1, – i, 1

$∴$ Total number cases $= 4 \times 4 = 16$

For $i^{k_{1}} + i^{k_{2}} = 0,$

Cases are (1, –1), (–1, 1), (i, – i), (– i, i)

$∴$ Required probability $= 1 - \frac{4}{16} = \frac{12}{16} = \frac{3}{4}$ .

Q 30 :

Bag $B_{1}$ contains 6 white and 4 blue balls, Bag $B_{2}$ contains 4 white and 6 blue balls, and Bag $B_{3}$ contains 5 white and 5 blue balls. One of the bags is selected at random and a ball is drawn from it. If the ball is white, then the probability, that the ball is drawn from bag $B_{2}$ , is : [2025]

$\frac{2}{5}$
$\frac{4}{15}$
$\frac{2}{3}$
$\frac{1}{3}$

1

2

3

4

(2)

Consider the events, $E_{1}$ : Bag $B_{1}$ is selected, $E_{2}$ : Bag $B_{2}$ is selected, $E_{3}$ : Bag $B_{3}$ is selected

A : Ball drawn is white. Then $P (E_{1}) = P (E_{2}) = P (E_{3}) = \frac{1}{3}$

$P (A / E_{1}) = \frac{6}{10}, P (A / E_{2}) = \frac{4}{10}, P (A / E_{3}) = \frac{5}{10}$

Required probability

$P (\frac{E_{2}}{A}) = \frac{P (E_{2}) P (\frac{A}{E_{2}})}{P (E_{1}) P (\frac{A}{E_{1}}) + P (E_{2}) P (\frac{A}{E_{2}}) + P (E_{3}) P (\frac{A}{E_{3}})}$

$= \frac{\frac{1}{3} \times \frac{4}{10}}{(\frac{1}{3}) (\frac{6}{10}) + (\frac{1}{3}) (\frac{4}{10}) + (\frac{1}{3}) (\frac{5}{10})} = \frac{\frac{4}{30}}{\frac{6}{30} + \frac{4}{30} + \frac{5}{30}} = \frac{4}{15}$

Topic Question Set

Q 1 :

Three urns A, B and C contain 7 red, 5 black, 5 red, 7 black and 6 red, 6 black balls, respectively. One of the urns is selected at random and a ball is drawn from it. If the ball drawn is black, then the probability that it is drawn from urn A is __________ . [2024]

Q 3 :

If three letters can be posted to any one of the 5 different addresses, then the probability that the three letters are posted to exactly two addresses is ______. [2024]

Q 4 :

Let the sum of two positive integers be 24. If the probability that their product is not less than $\frac{3}{4}$ times their greatest possible product is $\frac{m}{n},$ where $g c d (m, n) = 1,$ then $n - m$ equals [2024]

Q 6 :

If an unbiased dice is rolled thrice, then the probability of getting a greater number in the $i^{t h}$ roll than the number obtained in the ${(i - 1)}^{t h}$ roll, $i = 2, 3,$ is equal to [2024]

Q 7 :

A bag contains 8 balls, whose colours are either white or black. 4 balls are drawn at random without replacement and it was found that 2 balls are white and other 2 balls are black. The probability that the bag contains equal number of white and black balls is [2024]

Q 8 :

Let Ajay will not appear in JEE exam with probability $p = \frac{2}{7},$ while both Ajay and Vijay will appear in the exam with probability $q = \frac{1}{5} .$ Then the probability that Ajay will appear in the exam and Vijay will not appear is [2024]

Q 9 :

An urn contains 6 white and 9 black balls. Two successive draws of 4 balls are made without replacement. The probability that the first draw gives all white balls and the second draw gives all black balls, is [2024]

Q 10 :

A fair die is thrown until 2 appears. Then the probability that 2 appears in an even number of throws, is [2024]

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]

Q 21 :

A bag contains 19 unbiased coins and one coin with head on both sides. One coin drawn at random is tossed and head turns up. If the probability that the drawn coin was unbiased, is $\frac{m}{n}$ , gcd (m, n) = 1, then $n^{2} - m^{2}$ is equal to [2025]

Q 22 :

If A and B are two events such that P(A) = 0.7, P(B) = 0.4 and $P (A \cap \bar{B}) = 0.5$ , where $\bar{B}$ denotes the complement of B, then $P (B / (A \cup \bar{B}))$ is equal to [2025]

Q 24 :

If A and B are two events such that $P (A \cap B) = 0.1$ , and $P (A | B)$ and $P (B | A)$ are the roots of the equation $12 x^{2} - 7 x + 1 = 0$ , then the value of $\frac{P (\bar{A} \cup \bar{B})}{P (\bar{A} \cap \bar{B})}$ is : [2025]

Q 26 :

A board has 16 squares as shown in the figure:

Out of these 16 squares, two squares are chosen at random. The probability that they have no side in common is: [2025]

Q 27 :

A and B alternately throw a pair of dice. A wins if he throws a sum of 5 before B throws a sum of 8, and B wins if he throws a sum of 8 before A throws a sum of 5. The probability, that A wins if A makes the first throw, is [2025]

Q 28 :

Let $A = [a_{i j}]$ be a square matrix of order 2 with entries either 0 or 1. Let E be the event that A is an invertible matrix. Then the probability P(E) is: [2025]

Q 29 :

Two number $k_{1}$ and $k_{2}$ are randomly chosen from the set of natural numbers. Then, the probability that the value of $i^{k_{1}} + i^{k_{2}}, (i = \sqrt{- 1})$ is non-zero, equals [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 1 : Three urns A, B and C contain 7 red, 5 black, 5 red, 7 black and 6 red, 6 black balls, respectively. One of the urns is selected at random and a ball is drawn from it. If the ball drawn is black, then the probability that it is drawn from urn A is __________ . [2024]

Q 3 : If three letters can be posted to any one of the 5 different addresses, then the probability that the three letters are posted to exactly two addresses is ______. [2024]

Q 4 : Let the sum of two positive integers be 24. If the probability that their product is not less than 34 times their greatest possible product is mn, where gcd(m,n)=1, then n-m equals [2024]

Q 6 : If an unbiased dice is rolled thrice, then the probability of getting a greater number in the ith roll than the number obtained in the (i-1)th roll, i=2,3, is equal to [2024]

Q 7 : A bag contains 8 balls, whose colours are either white or black. 4 balls are drawn at random without replacement and it was found that 2 balls are white and other 2 balls are black. The probability that the bag contains equal number of white and black balls is [2024]

Q 8 : Let Ajay will not appear in JEE exam with probability p=27, while both Ajay and Vijay will appear in the exam with probability q=15. Then the probability that Ajay will appear in the exam and Vijay will not appear is [2024]

Q 9 : An urn contains 6 white and 9 black balls. Two successive draws of 4 balls are made without replacement. The probability that the first draw gives all white balls and the second draw gives all black balls, is [2024]

Q 10 : A fair die is thrown until 2 appears. Then the probability that 2 appears in an even number of throws, is [2024]

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) 13 K 16 14 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 ax2+bx+c=0 has all real roots is mn,gcd(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]

Q 21 : A bag contains 19 unbiased coins and one coin with head on both sides. One coin drawn at random is tossed and head turns up. If the probability that the drawn coin was unbiased, is mn, gcd (m, n) = 1, then n2–m2 is equal to [2025]

Q 22 : If A and B are two events such that P(A) = 0.7, P(B) = 0.4 and P(A∩B¯)=0.5, where B¯ denotes the complement of B, then P(B/(A∪B-)) is equal to [2025]

Q 23 : Two balls are selected at random one by one without replacement from a bag containing 4 white and 6 black balls. If the probability that the first selected balls is black, given that the second selected ball is also black is mn, where gcd (m, n) = 1, then m + n is equal to : [2025]

Q 24 : If A and B are two events such that P(A∩B)=0.1, and P(A|B) and P(B|A) are the roots of the equation 12x2–7x+1=0, then the value of P(A¯∪B¯)P(A¯∩B¯) is : [2025]

Q 26 : A board has 16 squares as shown in the figure: Out of these 16 squares, two squares are chosen at random. The probability that they have no side in common is: [2025]

Q 27 : A and B alternately throw a pair of dice. A wins if he throws a sum of 5 before B throws a sum of 8, and B wins if he throws a sum of 8 before A throws a sum of 5. The probability, that A wins if A makes the first throw, is [2025]

Q 28 : Let A=[aij] be a square matrix of order 2 with entries either 0 or 1. Let E be the event that A is an invertible matrix. Then the probability P(E) is: [2025]

Q 29 : Two number k1 and k2 are randomly chosen from the set of natural numbers. Then, the probability that the value of ik1+ik2, (i=–1) is non-zero, equals [2025]

Q 30 : Bag B1 contains 6 white and 4 blue balls, Bag B2 contains 4 white and 6 blue balls, and Bag B3 contains 5 white and 5 blue balls. One of the bags is selected at random and a ball is drawn from it. If the ball is white, then the probability, that the ball is drawn from bag B2, is : [2025]

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Q 1 :

Three urns A, B and C contain 7 red, 5 black, 5 red, 7 black and 6 red, 6 black balls, respectively. One of the urns is selected at random and a ball is drawn from it. If the ball drawn is black, then the probability that it is drawn from urn A is __________ . [2024]

Q 3 :

If three letters can be posted to any one of the 5 different addresses, then the probability that the three letters are posted to exactly two addresses is ______. [2024]

Q 4 :

Let the sum of two positive integers be 24. If the probability that their product is not less than $\frac{3}{4}$ times their greatest possible product is $\frac{m}{n},$ where $g c d (m, n) = 1,$ then $n - m$ equals [2024]

Q 6 :

If an unbiased dice is rolled thrice, then the probability of getting a greater number in the $i^{t h}$ roll than the number obtained in the ${(i - 1)}^{t h}$ roll, $i = 2, 3,$ is equal to [2024]

Q 7 :

A bag contains 8 balls, whose colours are either white or black. 4 balls are drawn at random without replacement and it was found that 2 balls are white and other 2 balls are black. The probability that the bag contains equal number of white and black balls is [2024]

Q 8 :

Let Ajay will not appear in JEE exam with probability $p = \frac{2}{7},$ while both Ajay and Vijay will appear in the exam with probability $q = \frac{1}{5} .$ Then the probability that Ajay will appear in the exam and Vijay will not appear is [2024]

Q 9 :

An urn contains 6 white and 9 black balls. Two successive draws of 4 balls are made without replacement. The probability that the first draw gives all white balls and the second draw gives all black balls, is [2024]

Q 10 :

A fair die is thrown until 2 appears. Then the probability that 2 appears in an even number of throws, is [2024]

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]

Q 21 :

A bag contains 19 unbiased coins and one coin with head on both sides. One coin drawn at random is tossed and head turns up. If the probability that the drawn coin was unbiased, is $\frac{m}{n}$ , gcd (m, n) = 1, then $n^{2} - m^{2}$ is equal to [2025]

Q 22 :

If A and B are two events such that P(A) = 0.7, P(B) = 0.4 and $P (A \cap \bar{B}) = 0.5$ , where $\bar{B}$ denotes the complement of B, then $P (B / (A \cup \bar{B}))$ is equal to [2025]

Q 24 :

If A and B are two events such that $P (A \cap B) = 0.1$ , and $P (A | B)$ and $P (B | A)$ are the roots of the equation $12 x^{2} - 7 x + 1 = 0$ , then the value of $\frac{P (\bar{A} \cup \bar{B})}{P (\bar{A} \cap \bar{B})}$ is : [2025]

Q 26 :

A board has 16 squares as shown in the figure:

Out of these 16 squares, two squares are chosen at random. The probability that they have no side in common is: [2025]

Q 27 :

A and B alternately throw a pair of dice. A wins if he throws a sum of 5 before B throws a sum of 8, and B wins if he throws a sum of 8 before A throws a sum of 5. The probability, that A wins if A makes the first throw, is [2025]

Q 28 :

Let $A = [a_{i j}]$ be a square matrix of order 2 with entries either 0 or 1. Let E be the event that A is an invertible matrix. Then the probability P(E) is: [2025]

Q 29 :

Two number $k_{1}$ and $k_{2}$ are randomly chosen from the set of natural numbers. Then, the probability that the value of $i^{k_{1}} + i^{k_{2}}, (i = \sqrt{- 1})$ is non-zero, equals [2025]