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.

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

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

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

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

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

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

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%

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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 mn, 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 1150, then n is equal to ___________. [2025]

0%

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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]

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]