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

Q 31 :

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 32 :

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]

0%

(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 33 :

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]

0%

(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 34 :

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]

0%

(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 35 :

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 36 :

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 37 :

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 38 :

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 39 :

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 40 :

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

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

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

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

Topic Question Set

Q 31 :

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 32 :

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]

0%

0%

Q 34 :

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]

0%

Q 35 :

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 37 :

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 38 :

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 39 :

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]

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