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Probability Part 2

Section: Mathematics

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Question : 50 of 100

Marks: +1, -0

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 \ge 2

A = \{2k+3\ell; k,\ell\in\mathbb{N}\}

B = \{ w_n; n \in A \}

P(B)

[29-Jan-2023 Shift 2]

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

Solution:

Let

P(w_1) = \lambda

then

P(w_2) = \frac{\lambda}{2} \dots P(w_n) = \frac{\lambda}{2^{n-1}}

As

\sum\limits_{k=1}^{\infty} P(w_k)

=1 \Rightarrow \frac{\lambda}{1-\frac{1}{2}} = 1 \Rightarrow \lambda = \frac{1}{2}

So,

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

A = \{2k+3\ell; k,\ell\in\mathbb{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 = \mathbb{N} - \{1,2,3,4,6\}

\therefore P(B) = 1 - [P(w_1) + P(w_2) + P(w_3) + P(w_4) + P(w_6)]

=1 - \left[\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{64}\right]

=1 - \frac{32+16+8+4+1}{64} = \frac{3}{64}

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