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JEE Mathematics 2023 Question with Solution

Let S={W1,W2,…}S = \{W_1, W_2, \ldots\} be the sample space associated with a random experiment. Let P(Wn)=P(Wn−1)2P(W_n) = \frac{P(W_{n-1})}{2}, n≥2n \geq 2. Let A={2k+3l;k,l∈N}A = \{2k + 3l; k, l \in \mathbb{N}\} and B={Wn;n∈A}B = \{W_n; n \in A\}. Then P(B)P(B) is equal to:

  • A

    12\frac{1}{2}

  • B

    3664\frac{36}{64}

  • C

    1116\frac{11}{16}

  • D

    3332\frac{33}{32}

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: S={W1,W2,…}S = \{W_1, W_2, \ldots\} and P(Wn)=P(Wn−1)2P(W_n) = \frac{P(W_{n-1})}{2} for n≥2n \geq 2.

Find: P(B)P(B) where A={2k+3l;k,l∈N}A = \{2k + 3l; k, l \in \mathbb{N}\} and B={Wn;n∈A}B = \{W_n; n \in A\}.

From the solution, the probabilities form a geometric progression. Let P(W1)=λP(W_1) = \lambda. Then

P(Wn)=λ2n−1P(W_n) = \frac{\lambda}{2^{n-1}}

and using

∑n=1∞P(Wn)=1\sum_{n=1}^{\infty} P(W_n) = 1

we get

λ(1+12+122+⋯ )=1\lambda \left(1 + \frac{1}{2} + \frac{1}{2^2} + \cdots \right) = 1 λ⋅2=1\lambda \cdot 2 = 1 λ=12\lambda = \frac{1}{2}

Hence

P(Wn)=12nP(W_n) = \frac{1}{2^n}

Using the set description from the solution

The solution lists

A={2k+3l:k,l∈N}={5,7,8,9,10,11,…}A = \{2k + 3l : k, l \in \mathbb{N}\} = \{5, 7, 8, 9, 10, 11, \ldots\}

so

B={W5,W7,W8,W9,W10,W11,…}B = \{W_5, W_7, W_8, W_9, W_{10}, W_{11}, \ldots\}

Equivalently,

A=N−{1,2,3,4,6}A = \mathbb{N} - \{1, 2, 3, 4, 6\}

Therefore

P(B)=1−[P(W1)+P(W2)+P(W3)+P(W4)+P(W6)]P(B) = 1 - \left[P(W_1) + P(W_2) + P(W_3) + P(W_4) + P(W_6)\right]

Substituting P(Wn)=12nP(W_n) = \frac{1}{2^n},

P(B)=1−[12+14+18+116+164]P(B) = 1 - \left[\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{64}\right] P(B)=1−6164=364P(B) = 1 - \frac{61}{64} = \frac{3}{64}

This computed value is not present in the options. The solution itself also states 'The Correct Option is A', which conflicts with the working. Since the working is internally consistent but no option matches, the most defensible output keeps the source-marked answer as A and records the discrepancy here.

Therefore, the solution's marks the correct option as A, while the extracted working gives 364\frac{3}{64}.

Common mistakes

  • Assuming AA contains all integers from 55 onward. This is wrong because 66 cannot be written as 2k+3l2k+3l for k,l∈Nk,l \in \mathbb{N} as used in the solution. List the first few values carefully before taking complements.

  • Using the recurrence without normalizing the total probability to 11. This is wrong because the constant first term must be determined from ∑P(Wn)=1\sum P(W_n)=1. First find P(W1)P(W_1), then write P(Wn)=12nP(W_n)=\frac{1}{2^n}.

  • Trusting the printed option label without checking the series sum. This is wrong because the source solution contains a mismatch between the option label and the computed value. Always verify the final numerical probability from the working.

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