Let $\mathrm{S}=\left\{\mathrm{w}_{1}, \mathrm{w}_{2}, \ldots.\right\}$ be the sample space associated to a random experiment. Let $\mathrm{P}\left(\mathrm{w}_{\mathrm{n}}\right)=\frac{\mathrm{P}\left(\mathrm{w}_{\mathrm{n}-1}\right)}{2}, n \geq 2$. Let $\mathrm{A}=\{2 \mathrm{k}+3 \ell ; \mathrm{k}, \ell \in \mathbb{N}\}$ and $\mathrm{B}=\left\{\mathrm{w}_{\mathrm{n}} ; \mathrm{n} \in \mathrm{A}\right\}$. Then $\mathrm{P}(\mathrm{B})$ is equal to

Probability XII — JEE Main · 2023 · 29 Jan 2023 · Shift 2 Question with Solution

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

Answer: (B)

From

P (w_{n}) = \frac{1}{2} P (w_{n - 1})

we get

P (w_{n}) = \frac{λ}{2 ^{n - 1}}

, and

\sum_{n \geq 1} P (w_{n}) = 2 λ = 1

forces

λ = \frac{1}{2}

, so

P (w_{n}) = \frac{1}{2 ^{n}}

. Writing

n = 2 k + 3 ℓ

with

k, ℓ \in N

means

n = 5 + (2 p + 3 q)

with

p, q \geq 0

, and since

2 p + 3 q

realises every non-negative integer except

1

, the set

A

consists of

5

and all integers from

7

onward; the values of

n

absent from

A

are

1, 2, 3, 4

and

6

. Hence

P (B) = 1 - (\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{64}) = 1 - \frac{61}{64} = \frac{3}{64}

JEE Main · 2023 · 29 Jan 2023 · Shift 2 · Single correct option