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

Q 11 :

Bag 1 contains 4 white balls and 5 black balls, and Bag 2 contains n white balls and 3 black balls. One ball is drawn randomly from Bag 1 and transferred to Bag 2. A ball is then drawn randomly from Bag 2. If the probability, that the ball drawn is white, 29/45, then n is equal to : [2025]

4
6
3
5

1

2

3

4

(2)

We have, Bag 1 : {4W, 5B} and Bag 2 : {nW, 3B}

Also, P(W/Bag 2) = $\frac{29}{45}$

$\Rightarrow P (\frac{W}{B_{1}}) \times P (\frac{W}{B_{2}}) + P (\frac{B}{B_{1}}) \times P (\frac{W}{B_{2}}) = \frac{29}{45}$

$\Rightarrow \frac{4}{9} (\frac{n + 1}{n + 4}) + \frac{5}{9} (\frac{n}{n + 4}) = \frac{29}{45} \Rightarrow \frac{4 n + 4 + 5 n}{9 n + 36} = \frac{29}{45}$

$\Rightarrow \frac{9 n + 4}{9 n + 36} = \frac{29}{45} \Rightarrow 405 n + 180 = 261 n + 1044$

$\Rightarrow 144 n = 864 \Rightarrow n = 6$ .

Q 12 :

Three distinct numbers are selected randomly from the set {1, 2, 3, ..., 40}. If the probability, that the selected numbers are in an increasing G.P., is $\frac{m}{n}$ , gcd (m, n) = 1, then m + n is __________. [2025]

(2477)

When $r \in N$

Common ratio	Last triplet	Total number of G.P. formed
r = 2	10, 20, 40	10
r = 3	4, 12, 36	4
r = 4	2, 8, 32	2
r = 5	1, 5, 25	1
r = 6	1, 6, 36	1
	Total	18

When $r \notin N$ (also possible)

Common ratio	Last triplet	Total number of G.P. formed
r = 3/2	16, 24, 36	4
r = 5/2	4, 10, 25	1
r = 4/3	18, 24, 32	2
r = 5/3	9, 15, 25	1
r = 5/4	16, 20, 25	1
r = 6/5	25, 30, 36	1
	Total	10

Total number of choices $= {}^{40}C_{3} = 9880$

Required probability $= \frac{28}{9880} = \frac{7}{2470} = \frac{m}{n}$

$∴$ m = 7 and n = 2470

Hence, m + n = 7 + 2470 = 2477.

Q 13 :

All five letter words are made using all the letters A, B, C, D, E and arranged as in an English dictionary with serial numbers. Let the word at serial number n be denoted by $W_{n}$ . Let the probability $P (W_{n})$ of choosing the word $W_{n}$ satisfy $P (W_{n}) = 2 P (W_{n - 1}), n > 1$ . If $P (C D B E A) = \frac{2^{α}}{2^{β} - 1}, α, β \in ℕ$ , then $α + β$ is equal to __________. [2025]

(183)

Let $P (W_{1}) = x$

Possible arrangement of 5 letters = 5! = 120

Now, $\sum_{i = 1}^{120} P (W_{i}) = 1$

$\Rightarrow x + 2 x + 2^{2} x + 2^{3} x + . . . . . + 2^{119} x = 1$

$\Rightarrow \frac{x (2^{120} - 1)}{(2 - 1)} = 1 \Rightarrow x = \frac{1}{2^{120} - 1}$ ... (1)

Possible arrangements after fixing letters are given by

A _ _ _ = 4! = 24

B _ _ _ _ = 4! = 24

C A _ _ _ = 3! = 6

C B _ _ _ = 3! = 6

C D A _ _ = 2! = 2

C D B A E = 1

C D B E A = 1

${Rank of C D B E A = 64}_{_________________________}^{_________________________}$

So, $P (W_{64}) = 2 P (W_{63}) = . . . = 2^{63} P (W_{1}) = \frac{2^{63}}{2^{120} - 1}$

On Comparing with $\frac{2^{α}}{2^{β} - 1}$ , we get $α$ = 63 and $β$ = 120.

$\Rightarrow α + β = 63 + 120 = 183$ .

Q 14 :

A card from a pack of 52 cards is lost. From the remaining 51 cards, n cards are drawn and are found to be spades. If the probability of the lost card to be a spade is $\frac{11}{50}$ , then n is equal to ___________. [2025]

(2)

Let us define the events:

A : Lost card is a spade

B : n cards drawn from 51 cards are spades.

$∴ P (A) = \frac{13}{52} = \frac{1}{4}, P (A^{c}) = 1 - \frac{1}{4} = \frac{3}{4}$

$P (B / A) = \frac{{}^{12}C_{n}}{{}^{51}C_{n}}, P (B / A^{c}) = \frac{{}^{13}C_{n}}{{}^{51}C_{n}}$

$P (B) = P (A) P (B / A) + P (A^{c}) P (B / A^{c})$

$= \frac{1}{4} \frac{{}^{12}C_{n}}{{}^{51}C_{n}} + \frac{3}{4} \frac{{}^{13}C_{n}}{{}^{51}C_{n}}$

$∴ P (\frac{A}{B}) = \frac{P (B / A) P (A)}{P (B)} \Rightarrow \frac{11}{50} = \frac{\frac{{}^{12}C_{n}}{{}^{51}C_{n}} \times \frac{1}{4}}{P (B)}$

$\Rightarrow \frac{13 - n}{52 - n} = \frac{11}{50} \Rightarrow n = 2$ .

Q 15 :

Three dice are rolled. If the probability of getting different numbers on the three dice is $\frac{p}{q},$ where $p$ and $q$ are co-prime, then $q - p$ is equal to [2023]

4
3
1
2

1

2

3

4

(1)

If numbers are different on all three dice,

Total number of favourable outcomes $= 6 \times 5 \times 4 = 120$

Total number of possible outcomes $= 6^{3} = 216$

$∴$ $Required probability = \frac{120}{216} = \frac{5}{9} = \frac{p}{q}$

So, $q - p = 9 - 5 = 4$

Q 16 :

In a bolt factory, machines A, B and C manufacture respectively 20%, 30% and 50% of the total bolts. Of their output 3%, 4%, and 2% are respectively defective bolts. A bolt is drawn at random from the product. If the bolt drawn is found defective, then the probability that it is manufactured by the machine C is [2023]

$\frac{3}{7}$
$\frac{9}{28}$
$\frac{5}{14}$
$\frac{2}{7}$

1

2

3

4

(3)

$P (A) = \frac{20}{100},$ $P (B) = \frac{30}{100},$ $P (C) = \frac{50}{100}$

$Let D be the event that the product is defective, then$

$P (D / A) = \frac{3}{100},$ $P (D / B) = \frac{4}{100},$ $P (D / C) = \frac{2}{100}$

$Now, by Bayes' theorem, we have$

$P (C / D) = \frac{P (C) \cdot P (D / C)}{P (A) \cdot P (D / A) + P (B) \cdot P (D / B) + P (C) \cdot P (D / C)}$

$= \frac{\frac{50}{100} \times \frac{2}{100}}{\frac{20}{100} \times \frac{3}{100} + \frac{30}{100} \times \frac{4}{100} + \frac{50}{100} \times \frac{2}{100}} = \frac{5}{14}$

Q 17 :

Let $N$ denote the sum of the numbers obtained when two dice are rolled. If the probability that $2^{N} < N!$ is $\frac{m}{n}$ , where $m$ and $n$ are coprime, then $4 m - 3 n$ is equal to [2023]

12
8
6
10

1

2

3

4

(2)

$2^{N} < N!, which is true when N \geq 4$

$N = 1 (Not possible)$

$N = 2 i.e., (1, 1) (Not possible)$

$N = 3 i.e., (1, 2), (2, 1) (Not possible)$

$∴$ $Required probability = \frac{36 - 3}{36} = \frac{33}{36} = \frac{11}{12}$

$∴$ $m = 11 and n = 12$

$Now, 4 m - 3 n = 4 (11) - 3 (12) = 44 - 36 = 8$

Q 18 :

Let $S = {M = [a_{i j}], a_{i j} \in {0, 1, 2}, 1 \leq i, j \leq 2}$ be a sample space and $A = {M \in S : M is invertible}$ be an event. Then $P (A)$ is equal to [2023]

$\frac{50}{81}$
$\frac{47}{81}$
$\frac{49}{81}$
$\frac{16}{27}$

1

2

3

4

(1)

$Let M = [\begin{matrix} p & q \\ r & s \end{matrix}]$ $where p, q, r, s \in {0, 1, 2}$

$n (S) = 3^{4} = 81$

$First, we find p (\bar{A})$

$\bar{A} = {M \in S : | M | = 0}$

$| M |$ $= 0 \Rightarrow p s = q r$

$p s = q r = 0 \Rightarrow number of (p, q, r, s) = {(3^{2} - 2^{2})}^{2} = 25$

$p s = q r = 1 \Rightarrow number of (p, q, r, s) = 1^{2} = 1$

$p s = q r = 2 \Rightarrow number of (p, q, r, s) = 2^{2} = 4$

$p s = q r = 4 \Rightarrow number of (p, q, r, s) = 1^{2} = 1$

$P (\bar{A}) = \frac{31}{81} \Rightarrow P (A) = \frac{50}{81}$

Q 19 :

A bag contains 6 white and 4 black balls. A die is rolled once, and the number of balls equal to the number obtained on the die are drawn from the bag at random. The probability that all the balls drawn are white is: [2023]

$\frac{9}{50}$
$\frac{1}{5}$
$\frac{11}{50}$
$\frac{1}{4}$

1

2

3

4

(2)

$Probability of getting any number on a die = \frac{1}{6}$

$Probability of getting all white balls when 1 appears on die = \frac{1}{6} \times \frac{{}^{6}{C_{1}}}{{}^{10}{C_{1}}} = \frac{1}{6} \times \frac{6}{10} = \frac{1}{10}$

$Probability of getting all white balls when 2 appears on die = \frac{1}{6} \times \frac{{}^{6}{C_{2}}}{{}^{10}{C_{2}}} = \frac{1}{6} \times \frac{6 \times 5}{10 \times 9} = \frac{1}{18}$

$Probability of getting all white balls when 3 appears on die = \frac{1}{6} \times \frac{{}^{6}{C_{3}}}{{}^{10}{C_{3}}} = \frac{1}{6} \times \frac{6 \times 5 \times 4}{10 \times 9 \times 8} = \frac{1}{36}$

$Probability of getting all white balls when 4 appears on die = \frac{1}{84}$

$Probability of getting all white balls when 5 appears on die = \frac{1}{252}$

$Probability of getting all white balls when 6 appears on die = \frac{1}{1260}$

$Required Probability = \frac{1}{10} + \frac{1}{18} + \frac{1}{36} + \frac{1}{84} + \frac{1}{252} + \frac{1}{1260} = \frac{1}{5}$

Q 20 :

Two dice are thrown independently. Let A be the event that the number appeared on the 1st die is less than the number appeared on the 2nd die, B be the event that the number appeared on the 1st die is even and that on the 2nd die is odd, and C be the event that the number appeared on the 1st die is odd and that on the 2nd is even. Then [2023]

A and B are mutually exclusive
The number of favourable cases of the events A, B and C are 15, 6 and 6 respectively.
B and C are independent
The number of favourable cases of the event $(A \cup B) \cap C$ is 6

1

2

3

4

(4)

Event A : Number on 1st die < Number on 2nd die

Event B : Number on 1st die = even and number on 2nd die = odd

Event C : Number on 1st die = odd and number on 2nd die = even

$n (A) = 5 + 4 + 3 + 2 + 1 = 15$

$n (B) = 9,$ $n (C) = 9$

$n ((A \cup B) \cap C) = (A \cap C) \cup (B \cap C) = (3 + 2 + 1) + 0 = 6$

Q 21 :

Let $M$ be the maximum value of the product of two positive integers when their sum is 66. Let the sample space $S = {x \in Z : x (66 - x) \geq \frac{5}{9} M}$ and the event $A = {x \in S : x is a multiple of 3} .$ Then $P (A)$ is equal to [2023]

$\frac{15}{44}$
$\frac{1}{3}$
$\frac{7}{22}$
$\frac{1}{5}$

1

2

3

4

(2)

$M = 33^{2} (Using A.M. and G.M. inequality)$

$x (66 - x) \geq \frac{5}{9} \times 33^{2}$

$\Rightarrow x^{2} - 66 x + \frac{5}{9} \times 33^{2} \leq 0 \Rightarrow x^{2} - 66 x + 605 \leq 0$ ...(i)

$\Rightarrow (x - 55) (x - 11) \leq 0;$ $11 \leq x \leq 55$

$∴$ $S = {11, 12, 13, \dots, 55} \Rightarrow n (S) = 45$

$Elements of S which are multiples of 3 are$

$12 + (n - 1) 3 = 54 \Rightarrow 3 (n - 1) = 42 \Rightarrow n = 15$

$n (A) = 15 \Rightarrow P (A) = \frac{15}{45} = \frac{1}{3}$

Q 22 :

Let $N$ be the sum of the numbers appeared when two fair dice are rolled and let the probability that $N - 2,$ $\sqrt{3 N},$ $N + 2$ are in geometric progression be $\frac{k}{48} .$ Then the value of $k$ is [2023]

16
2
8
4

1

2

3

4

(4)

$Here, n (S) = 36$

$N denotes the sum of numbers when two fair dice are rolled.$

$N - 2, \sqrt{3 N}, N + 2 are in G.P.$

$3 N = (N - 2) (N + 2)$

$\Rightarrow 3 N = N^{2} - 4 \Rightarrow N^{2} - 3 N - 4 = 0$

$\Rightarrow (N - 4) (N + 1) = 0$

$\Rightarrow N = 4$ $(since N = - 1 rejected)$

$Favorable cases: (1, 3), (3, 1), (2, 2)$

$Required probability = \frac{3}{36} = \frac{k}{48} (given) \Rightarrow k = 4$

Q 23 :

Let $S = {w_{1}, w_{2}, \dots}$ be the sample space associated to a random experiment. Let $P (w_{n}) = \frac{P (w_{n - 1})}{2}, n \geq 2 .$ Let $A = {2 k + 3 l : k, l \in ℕ}$ and $B = {w_{n} : n \in A}$ . Then $P (B)$ is equal to [2023]

$\frac{1}{16}$
$\frac{3}{64}$
$\frac{3}{32}$
$\frac{1}{32}$

1

2

3

4

(2)

Given $S = {w_{1}, w_{2}, \dots}$ be the sample space associated to a random experiment. If

$P (W_{n}) = \frac{P (W_{n - 1})}{2}, n \geq 2$

and let $A = {2 k + 3 l : k, l \in N}$ and $B = {w_{n} : n \in A}$ .

First of all, let $P (w_{1}) = λ$ . Then $P (w_{2}) = \frac{λ}{2}$

$P (w_{n}) = \frac{λ}{2^{n - 1}}$

As
$\sum_{k = 1}^{\infty} P (w_{k}) = 1 \Rightarrow \frac{λ}{1 - \frac{1}{2}} = 1 \Rightarrow λ = \frac{1}{2}$

So $P (W_{n}) = \frac{1}{2^{n}}$

$A = {2 k + 3 l : k, l \in N} = {5, 7, 8, 9, 10, \dots}$

$B = {w_{n} : n \in A}$

$B = {w_{5}, w_{7}, w_{8}, w_{9}, w_{10}, w_{11}, \dots}$

$A = N - {1, 2, 3, 4, 6}$

$P (B) = 1 - [P (w_{1}) + P (w_{2}) + P (w_{3}) + P (w_{4}) + P (w_{6})]$

$= 1 - [\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{64}] = 1 - \frac{32 + 16 + 8 + 4 + 1}{64} = \frac{3}{64}$

Q 24 :

A bag contains 6 balls. Two balls are drawn from it at random and both are found to be black. The probability that the bag contains at least 5 black balls is [2023]

$\frac{2}{7}$
$\frac{3}{7}$
$\frac{5}{7}$
$\frac{5}{6}$

1

2

3

4

(3)

Total possibilities :

Case I : 2B + 4 others

Case II : 3B + 3 others

Case III : 4B + 2 others

Case IV : 5B + 1 other

Case V : 6B + 0 other

$Required probability = \frac{\frac{{}^{5}{C_{2}}}{{}^{6}{C_{2}}} + \frac{{}^{6}{C_{2}}}{{}^{6}{C_{2}}}}{\frac{{}^{2}{C_{2}}}{{}^{6}{C_{2}}} + \frac{{}^{3}{C_{2}}}{{}^{6}{C_{2}}} + \frac{{}^{4}{C_{2}}}{{}^{6}{C_{2}}} + \frac{{}^{5}{C_{2}}}{{}^{6}{C_{2}}} + \frac{{}^{6}{C_{2}}}{{}^{6}{C_{2}}}}$

$= \frac{10 + 15}{1 + 3 + 6 + 10 + 15} = \frac{25}{35} = \frac{5}{7}$

Q 25 :

Let the probability of getting head for a biased coin be $\frac{1}{4}$ . It is tossed repeatedly until a head appears. Let $N$ be the number of tosses required. If the probability that the equation $64 x^{2} + 5 N x + 1 = 0$ has no real root is $\frac{p}{q}$ , where $p$ and $q$ are co-prime, then $q - p$ is equal to ______. [2023]

(27)

Let $H$ be the event of getting head and $T$ be the event of getting tail. $P (H) = \frac{1}{4} and P (T) = \frac{3}{4}$

Now, $64 x^{2} + 5 N x + 1 = 0$

$D = 25 N^{2} - 4 (64) (1) = 25 N^{2} - 256$

For no real roots, $D < 0$

$\Rightarrow 25 N^{2} - 256 < 0$

$\Rightarrow 25 N^{2} < 256 \Rightarrow N^{2} < \frac{256}{25} \Rightarrow N < \frac{16}{5}$

$∴$ $N = 1, 2, 3$

Required probability $= H + T H + T T H$

$= \frac{1}{4} + \frac{3}{4} \times \frac{1}{4} + \frac{3}{4} \times \frac{3}{4} \times \frac{1}{4}$

$= \frac{1}{4} + \frac{3}{16} + \frac{9}{64} = \frac{37}{64} = \frac{p}{q} (Given)$

$∴$ $q - p = 64 - 37 = 27$

Q 26 :

Three urns A, B and C contain 4 red, 6 black; 5 red, 5 black; and $λ$ red, 4 black balls respectively. One of the urns is selected at random and a ball is drawn. If the ball drawn is red and the probability that it is drawn from urn C is 0.4, then the square of the length of the side of the largest equilateral triangle, inscribed in the parabola $y^{2} = λ x$ with one vertex at the vertex of the parabola, is __________. [2023]

(432)

$P (Red ball from urn C)$

$=$ $\frac{\frac{1}{3} \cdot \frac{λ}{λ + 4}}{\frac{1}{3} \cdot \frac{4}{10} + \frac{1}{3} \cdot \frac{5}{10} + \frac{1}{3} \cdot \frac{λ}{λ + 4}} = 0.4 \Rightarrow \frac{\frac{λ}{λ + 4}}{\frac{4}{10} + \frac{5}{10} + \frac{λ}{λ + 4}} = \frac{4}{10}$

$∴$ $λ = 6$

So, parabola $y^{2} = 6 x$

Let side length of the triangle be $l$ .

$\tan 30 ° = \frac{3 t}{\frac{3}{2} t^{2}}$

$\Rightarrow \frac{1}{\sqrt{3}} = \frac{2}{t}$

$∴$ $t = 2 \sqrt{3}$

So, $(\frac{3}{2} t^{2}, 3 t) = (18, 6 \sqrt{3})$

Now, $l^{2} = 18^{2} + {(6 \sqrt{3})}^{2} = 324 + 108 = 432$

Q 27 :

25% of the population are smokers. A smoker has 27 times more chances to develop lung cancer than a non-smoker. A person is diagnosed with lung cancer and the probability that this person is a smoker is $\frac{k}{10}$ . Then the value of $k$ is _______. [2023]

(9)

Let number of smokers be $E_{1}$ . So, $P (E_{1}) = \frac{1}{4}$ .

The number of non-smokers be $E_{2}$ . So, $P (E_{2}) = \frac{3}{4}$ .

Let E denote persons diagnosed with lung cancer.

Let $P (E / E_{2}) = x, P (E / E_{1}) = 27 x$

Now, $P (E_{1} / E) = \frac{P (E_{1}) P (E / E_{1})}{P (E)}$

$= \frac{\frac{1}{4} \times 27 \times x}{\frac{1}{4} \times 27 \times x + \frac{3}{4} \times x} = \frac{27}{30} = \frac{9}{10}$

So, $k = 9$

Q 28 :

A bag contains six balls of different colours. Two balls are drawn in succession with replacement. The probability that both the balls are of the same colour is $p .$ Next four balls are drawn in succession with replacement and the probability that exactly three balls are of the same colour is $q$ . If $p : q = m : n$ ,where $m$ and $n$ are coprime, then $m + n$ is equal to _______. [2023]

(14)

Bag contains six balls of different colours.

Probability of drawing a ball of one colour $= \frac{1}{6}$

$∴$ $Probability of drawing two balls of same colour with replacement = p = {(\frac{1}{6})}^{2}$

Probability that exactly three balls are of same colour when four balls are drawn in succession with replacement.

$= q = 4 {(\frac{1}{6})}^{3} \times \frac{5}{6} = \frac{20}{6^{4}}$

$p : q = \frac{1}{6^{2}} : \frac{20}{6^{4}} = \frac{36}{20} = \frac{9}{5}$

$p : q = m : n = 9 : 5$ $∴$ $m = 9, n = 5$

$Now, m + n = 9 + 5 = 14$

Q 29 :

Let A be the event that the absolute difference between two randomly chosen real numbers in the sample space [0, 60] is less than or equal to $a$ .If $P (A) = \frac{11}{36},$ then $a$ is equal to ______ . [2023]

(10)

$| x - y | \leq a$

$\Rightarrow - a \leq x - y \leq a \Rightarrow x - y \leq a and x - y \geq - a$

$P (A) = \frac{ar (O A C D E G)}{ar (O B D F)} = \frac{ar (O B D F) - ar (∆ A B C) - ar (∆ E F G)}{ar (O B D F)}$

$\Rightarrow \frac{11}{36} = \frac{60^{2} - \frac{1}{2} {(60 - a)}^{2} - \frac{1}{2} {(60 - a)}^{2}}{3600}$

$\Rightarrow 1100 = 3600 - {(60 - a)}^{2}$

$\Rightarrow {(60 - a)}^{2} = 2500 \Rightarrow 60 - a = 50 \Rightarrow a = 10$

Q 30 :

From the first 100 natural numbers, two numbers first a and then b are selected randomly without replacement. If the probability that $a - b \geq 10$ is $\frac{m}{n}$ , $\gcd (m, n) = 1,$ then $m + n$ is equal to _______ . [2026]

(311)

$a - b \geq 10$

$Total cases = 100 \times 99$

$Fav. cases = 1 + 2 + 3 + \dots + 90$

$Req. Prob = \frac{1 + 2 + \dots + 90}{100 \times 99}$

$\frac{m}{n} = \frac{90 (\frac{91}{2})}{100 (99)} = \frac{91}{220}$

$m + n = 311$

Q 31 :

From a lot containing 10 defective and 90 non-defective bulbs, 8 bulbs are selected one by one with replacement. Then the probability of getting at least 7 defective bulbs is [2026]

$\frac{73}{10^{8}}$
$\frac{67}{10^{8}}$
$\frac{81}{10^{8}}$
$\frac{7}{10^{7}}$

1

2

3

4

(1)

$10 defective & 90 non-defective$

$Req. probability = (7 defective & 1 fair) or (8 defective)$

$Req. probability = \frac{(10^{7} \times 90) \times 8 + 10^{8}}{100^{8}}$

$= \frac{72 \times 10^{8} + 10^{8}}{100^{8}} = \frac{73}{10^{8}}$

Q 32 :

Two distinct numbers $a$ and $b$ are selected at random from $1, 2, 3, \dots, 50$ . The probability that their product $a b$ is divisible by 3, is [2026]

$\frac{664}{1225}$
$\frac{561}{1225}$
$\frac{8}{25}$
$\frac{272}{1225}$

1

2

3

4

(1)

$Required probability = 1 - (product not divisible by 3)$

$Multiple of 3 = 16$

$Not multiple of 3 = 34$

$= 1 - \frac{{}^{34}C_{2}}{{}^{50}C_{2}}$

$= \frac{664}{1225}$

Q 33 :

Let $S$ be a set of 5 elements and P(S) denote the power set of S. Let E be an event of choosing an ordered pair (A, B) from the set $P (S) \times P (S)$ such that $A \cap B = \emptyset$ . If the probability of the event E is $\frac{3^{p}}{2^{q}},$ where $p, q \in ℕ$ , then $p + q$ is equal to __________ . [2026]

(15)

$S = {a, b, c, d, e}$

$P (S) contains 2^{5} = 32 elements$

$both set A and set B are subsets of P (S)$

$Every element has 4 choices$

$Favourable cases = 3^{5}$

$Total cases = 4^{5}$

$P = \frac{3^{5}}{4^{5}} = \frac{3^{5}}{2^{10}}$

$m = 5, n = 10$

$m + n = 15$

Q 34 :

Bag A contains 9 white and 8 black balls, while bag B contains 6 white and 4 black balls. One ball is randomly picked up from the bag B and mixed up with the balls in the bag A. Then a ball is randomly drawn from the bag A. If the probability, that the ball drawn is white, is $\frac{p}{q}$ , gcd (p,q) = 1, then $p + q$ is equal to [2026]

23
24
21
22

1

2

3

4

(1)

$∴$ $P (Drawn ball is white) = \frac{3}{5} \times \frac{10}{18} + \frac{2}{5} \times \frac{9}{18}$

$= \frac{48}{90} = \frac{8}{15} = \frac{p}{q}$

$∴$ $p + q = 23$

Q 35 :

Let n be the number obtained on rolling a fair die. If the probability that the system

$\begin{matrix} x - n y + z = 6 \\ x + (n - 2) y + (n + 1) z = 8 \\ (n - 1) y + z = 1 \end{matrix}$

Has a unique solution is $\frac{k}{6}$ , then the sum of k and all possible values of n is: [2026]

20
24
21
22

1

2

3

4

(4)

$x - n y + z = 6$

$x + (n - 2) y + (n + 1) z = 8$

$(n - 1) y + z = 1$

$| \begin{matrix} 1 & - n & 1 \\ 1 & (n - 2) & n + 1 \\ 0 & n - 1 & 1 \end{matrix} |$ $\neq 0$

$\Rightarrow n^{2} - 3 n + 2 \neq 0$

$n \neq 1, 2$

$For unique solution n = 3, 4, 5, 6$

$Now P (probability when system of equations has unique solution) = \frac{4}{6}$

$So k = 4$

$Now required sum = 4 + (3 + 4 + 5 + 6) = 22$

Q 36 :

A bag contains 10 balls out of which k are red and (10−k) are black, where $0 \leq k \leq 10$ . If three balls are drawn at random without replacement and all of them are found to be black, then the probability that the bag contains 1 red and 9 black balls is: [2026]

$\frac{7}{110}$
$\frac{7}{55}$
$\frac{7}{11}$
$\frac{14}{55}$

1

2

3

4

(4)

$Probability = \frac{{}^{1}C_{0} \cdot {}^{9}C_{3}}{\sum_{k = 0}^{10} {}^{k}C_{0} \cdot {}^{10 - k}C_{3}}$

$= \frac{{}^{9}C_{3}}{{}^{10}C_{3} + {}^{9}C_{3} + {}^{8}C_{3} + \dots + {}^{3}C_{3}}$

$= \frac{{}^{9}C_{3}}{{}^{11}C_{4}}$

$= \frac{14}{55}$

Topic Question Set

Q 11 :

Bag 1 contains 4 white balls and 5 black balls, and Bag 2 contains n white balls and 3 black balls. One ball is drawn randomly from Bag 1 and transferred to Bag 2. A ball is then drawn randomly from Bag 2. If the probability, that the ball drawn is white, 29/45, then n is equal to : [2025]

Q 12 :

Three distinct numbers are selected randomly from the set {1, 2, 3, ..., 40}. If the probability, that the selected numbers are in an increasing G.P., is $\frac{m}{n}$ , gcd (m, n) = 1, then m + n is __________. [2025]

Q 14 :

A card from a pack of 52 cards is lost. From the remaining 51 cards, n cards are drawn and are found to be spades. If the probability of the lost card to be a spade is $\frac{11}{50}$ , then n is equal to ___________. [2025]

Q 15 :

Three dice are rolled. If the probability of getting different numbers on the three dice is $\frac{p}{q},$ where $p$ and $q$ are co-prime, then $q - p$ is equal to [2023]

Q 17 :

Let $N$ denote the sum of the numbers obtained when two dice are rolled. If the probability that $2^{N} < N!$ is $\frac{m}{n}$ , where $m$ and $n$ are coprime, then $4 m - 3 n$ is equal to [2023]

Q 18 :

Let $S = {M = [a_{i j}], a_{i j} \in {0, 1, 2}, 1 \leq i, j \leq 2}$ be a sample space and $A = {M \in S : M is invertible}$ be an event. Then $P (A)$ is equal to [2023]

Q 19 :

A bag contains 6 white and 4 black balls. A die is rolled once, and the number of balls equal to the number obtained on the die are drawn from the bag at random. The probability that all the balls drawn are white is: [2023]

Q 21 :

Let $M$ be the maximum value of the product of two positive integers when their sum is 66. Let the sample space $S = {x \in Z : x (66 - x) \geq \frac{5}{9} M}$ and the event $A = {x \in S : x is a multiple of 3} .$ Then $P (A)$ is equal to [2023]

Q 22 :

Let $N$ be the sum of the numbers appeared when two fair dice are rolled and let the probability that $N - 2,$ $\sqrt{3 N},$ $N + 2$ are in geometric progression be $\frac{k}{48} .$ Then the value of $k$ is [2023]

Q 23 :

Let $S = {w_{1}, w_{2}, \dots}$ be the sample space associated to a random experiment. Let $P (w_{n}) = \frac{P (w_{n - 1})}{2}, n \geq 2 .$ Let $A = {2 k + 3 l : k, l \in ℕ}$ and $B = {w_{n} : n \in A}$ . Then $P (B)$ is equal to [2023]

Q 24 :

A bag contains 6 balls. Two balls are drawn from it at random and both are found to be black. The probability that the bag contains at least 5 black balls is [2023]

Q 27 :

25% of the population are smokers. A smoker has 27 times more chances to develop lung cancer than a non-smoker. A person is diagnosed with lung cancer and the probability that this person is a smoker is $\frac{k}{10}$ . Then the value of $k$ is _______. [2023]

Q 29 :

Let A be the event that the absolute difference between two randomly chosen real numbers in the sample space [0, 60] is less than or equal to $a$ .If $P (A) = \frac{11}{36},$ then $a$ is equal to ______ . [2023]

Q 30 :

From the first 100 natural numbers, two numbers first a and then b are selected randomly without replacement. If the probability that $a - b \geq 10$ is $\frac{m}{n}$ , $\gcd (m, n) = 1,$ then $m + n$ is equal to _______ . [2026]

Q 31 :

From a lot containing 10 defective and 90 non-defective bulbs, 8 bulbs are selected one by one with replacement. Then the probability of getting at least 7 defective bulbs is [2026]

Q 32 :

Two distinct numbers $a$ and $b$ are selected at random from $1, 2, 3, \dots, 50$ . The probability that their product $a b$ is divisible by 3, is [2026]

Q 35 :

Let n be the number obtained on rolling a fair die. If the probability that the system

$\begin{matrix} x - n y + z = 6 \\ x + (n - 2) y + (n + 1) z = 8 \\ (n - 1) y + z = 1 \end{matrix}$

Has a unique solution is $\frac{k}{6}$ , then the sum of k and all possible values of n is: [2026]

Q 36 :

A bag contains 10 balls out of which k are red and (10−k) are black, where $0 \leq k \leq 10$ . If three balls are drawn at random without replacement and all of them are found to be black, then the probability that the bag contains 1 red and 9 black balls is: [2026]

Want Us to Email you About Special Offers & Updates?

Topic Question Set

Q 11 : Bag 1 contains 4 white balls and 5 black balls, and Bag 2 contains n white balls and 3 black balls. One ball is drawn randomly from Bag 1 and transferred to Bag 2. A ball is then drawn randomly from Bag 2. If the probability, that the ball drawn is white, 29/45, then n is equal to : [2025]

Q 12 : Three distinct numbers are selected randomly from the set {1, 2, 3, ..., 40}. If the probability, that the selected numbers are in an increasing G.P., is mn, gcd (m, n) = 1, then m + n is __________. [2025]

Q 14 : A card from a pack of 52 cards is lost. From the remaining 51 cards, n cards are drawn and are found to be spades. If the probability of the lost card to be a spade is 1150, then n is equal to ___________. [2025]

Q 15 : Three dice are rolled. If the probability of getting different numbers on the three dice is pq, where p and q are co-prime, then q−p is equal to [2023]

Q 17 : Let N denote the sum of the numbers obtained when two dice are rolled. If the probability that 2N<N! is mn, where m and n are coprime, then 4m-3n is equal to [2023]

Q 18 : Let S={M=[aij],aij∈{0,1,2},1≤i,j≤2} be a sample space and A={M∈S:M is invertible} be an event. Then P(A) is equal to [2023]

Q 19 : A bag contains 6 white and 4 black balls. A die is rolled once, and the number of balls equal to the number obtained on the die are drawn from the bag at random. The probability that all the balls drawn are white is: [2023]

Q 21 : Let M be the maximum value of the product of two positive integers when their sum is 66. Let the sample space S={x∈Z:x(66-x)≥59M} and the event A={x∈S:x is a multiple of 3}. Then P(A) is equal to [2023]

Q 22 : Let N be the sum of the numbers appeared when two fair dice are rolled and let the probability that N-2, 3N, N+2 are in geometric progression be k48. Then the value of k is [2023]

Q 23 : Let S={w1,w2,…} be the sample space associated to a random experiment. Let P(wn)=P(wn-1)2,n≥2. Let A={2k+3l:k,l∈ℕ} and B={wn:n∈A}. Then P(B) is equal to [2023]

Q 24 : A bag contains 6 balls. Two balls are drawn from it at random and both are found to be black. The probability that the bag contains at least 5 black balls is [2023]

Q 25 : Let the probability of getting head for a biased coin be 14. It is tossed repeatedly until a head appears. Let N be the number of tosses required. If the probability that the equation 64x2+5Nx+1=0 has no real root is pq, where p and q are co-prime, then q-p is equal to ______. [2023]

Q 27 : 25% of the population are smokers. A smoker has 27 times more chances to develop lung cancer than a non-smoker. A person is diagnosed with lung cancer and the probability that this person is a smoker is k10. Then the value of k is _______. [2023]

Q 29 : Let A be the event that the absolute difference between two randomly chosen real numbers in the sample space [0, 60] is less than or equal to a.If P(A)=1136, then a is equal to ______ . [2023]

Q 30 : From the first 100 natural numbers, two numbers first a and then b are selected randomly without replacement. If the probability that a-b≥10 is mn, gcd(m,n)=1, then m+n is equal to _______ . [2026]

Q 31 : From a lot containing 10 defective and 90 non-defective bulbs, 8 bulbs are selected one by one with replacement. Then the probability of getting at least 7 defective bulbs is [2026]

Q 32 : Two distinct numbers a and b are selected at random from 1,2,3,…,50. The probability that their product ab is divisible by 3, is [2026]

Q 33 : Let S be a set of 5 elements and P(S) denote the power set of S. Let E be an event of choosing an ordered pair (A, B) from the set P(S)×P(S) such that A∩B=∅. If the probability of the event E is 3p2q, where p,q∈ℕ, then p+q is equal to __________ . [2026]

Q 35 : Let n be the number obtained on rolling a fair die. If the probability that the system x-ny+z=6x+(n-2)y+(n+1)z=8(n-1)y+z=1 Has a unique solution is k6, then the sum of k and all possible values of n is: [2026]

Q 36 : A bag contains 10 balls out of which k are red and (10−k) are black, where 0≤k≤10. If three balls are drawn at random without replacement and all of them are found to be black, then the probability that the bag contains 1 red and 9 black balls is: [2026]

Want Us to Email you About Special Offers & Updates?

Q 11 :

Bag 1 contains 4 white balls and 5 black balls, and Bag 2 contains n white balls and 3 black balls. One ball is drawn randomly from Bag 1 and transferred to Bag 2. A ball is then drawn randomly from Bag 2. If the probability, that the ball drawn is white, 29/45, then n is equal to : [2025]

Q 12 :

Three distinct numbers are selected randomly from the set {1, 2, 3, ..., 40}. If the probability, that the selected numbers are in an increasing G.P., is $\frac{m}{n}$ , gcd (m, n) = 1, then m + n is __________. [2025]

Q 14 :

A card from a pack of 52 cards is lost. From the remaining 51 cards, n cards are drawn and are found to be spades. If the probability of the lost card to be a spade is $\frac{11}{50}$ , then n is equal to ___________. [2025]

Q 15 :

Three dice are rolled. If the probability of getting different numbers on the three dice is $\frac{p}{q},$ where $p$ and $q$ are co-prime, then $q - p$ is equal to [2023]

Q 17 :

Let $N$ denote the sum of the numbers obtained when two dice are rolled. If the probability that $2^{N} < N!$ is $\frac{m}{n}$ , where $m$ and $n$ are coprime, then $4 m - 3 n$ is equal to [2023]

Q 18 :

Let $S = {M = [a_{i j}], a_{i j} \in {0, 1, 2}, 1 \leq i, j \leq 2}$ be a sample space and $A = {M \in S : M is invertible}$ be an event. Then $P (A)$ is equal to [2023]

Q 19 :

A bag contains 6 white and 4 black balls. A die is rolled once, and the number of balls equal to the number obtained on the die are drawn from the bag at random. The probability that all the balls drawn are white is: [2023]

Q 21 :

Let $M$ be the maximum value of the product of two positive integers when their sum is 66. Let the sample space $S = {x \in Z : x (66 - x) \geq \frac{5}{9} M}$ and the event $A = {x \in S : x is a multiple of 3} .$ Then $P (A)$ is equal to [2023]

Q 22 :

Let $N$ be the sum of the numbers appeared when two fair dice are rolled and let the probability that $N - 2,$ $\sqrt{3 N},$ $N + 2$ are in geometric progression be $\frac{k}{48} .$ Then the value of $k$ is [2023]

Q 23 :

Let $S = {w_{1}, w_{2}, \dots}$ be the sample space associated to a random experiment. Let $P (w_{n}) = \frac{P (w_{n - 1})}{2}, n \geq 2 .$ Let $A = {2 k + 3 l : k, l \in ℕ}$ and $B = {w_{n} : n \in A}$ . Then $P (B)$ is equal to [2023]

Q 24 :

A bag contains 6 balls. Two balls are drawn from it at random and both are found to be black. The probability that the bag contains at least 5 black balls is [2023]

Q 27 :

25% of the population are smokers. A smoker has 27 times more chances to develop lung cancer than a non-smoker. A person is diagnosed with lung cancer and the probability that this person is a smoker is $\frac{k}{10}$ . Then the value of $k$ is _______. [2023]

Q 29 :

Let A be the event that the absolute difference between two randomly chosen real numbers in the sample space [0, 60] is less than or equal to $a$ .If $P (A) = \frac{11}{36},$ then $a$ is equal to ______ . [2023]

Q 30 :

From the first 100 natural numbers, two numbers first a and then b are selected randomly without replacement. If the probability that $a - b \geq 10$ is $\frac{m}{n}$ , $\gcd (m, n) = 1,$ then $m + n$ is equal to _______ . [2026]

Q 31 :

From a lot containing 10 defective and 90 non-defective bulbs, 8 bulbs are selected one by one with replacement. Then the probability of getting at least 7 defective bulbs is [2026]

Q 32 :

Two distinct numbers $a$ and $b$ are selected at random from $1, 2, 3, \dots, 50$ . The probability that their product $a b$ is divisible by 3, is [2026]

Q 35 :

Let n be the number obtained on rolling a fair die. If the probability that the system

$\begin{matrix} x - n y + z = 6 \\ x + (n - 2) y + (n + 1) z = 8 \\ (n - 1) y + z = 1 \end{matrix}$

Has a unique solution is $\frac{k}{6}$ , then the sum of k and all possible values of n is: [2026]

Q 36 :

A bag contains 10 balls out of which k are red and (10−k) are black, where $0 \leq k \leq 10$ . If three balls are drawn at random without replacement and all of them are found to be black, then the probability that the bag contains 1 red and 9 black balls is: [2026]