Probability Question 8
Question 8 - 29 January - Shift 2
Let $S={w_1, w_2, \ldots.}$ be the sample space associated to a random experiment. Let $P(w_n)=\frac{P(w _{n-1})}{2}, n \geq 2$.Let $A={2 k+3 \ell ; k, \ell \in \mathbb{N}}$ and $B={w_n ; n \in A}$.Then $P(B)$ is equal to
(1) $\frac{3}{32}$
(2) $\frac{3}{64}$
(3) $\frac{1}{16}$
(4) $\frac{1}{32}$
Show Answer
Answer: (2)
Solution:
Formula: Probability of occurrence of an event, Probability of a sure event, Important results on probability
Let $P(w_1)=\lambda$ then $P(w_2)=\frac{\lambda}{2} \ldots P(w_n)=\frac{\lambda}{2^{n-1}}$
As $\sum _{k=1}^{\infty} P(w_k)=1 \Rightarrow \frac{\lambda}{1-\frac{1}{2}}=1 \Rightarrow \lambda=\frac{1}{2}$
So, $P(w_n)=\frac{1}{2^{n}}$
$A={2 k+3 \ell ; k, \ell \in \mathbb{N}}={5,7,8,9,10 \ldots .$.
$B={w_n: n \in A}$
$B={w_5, w_7, w_8, w_9, w _{10}, w _{11}, \ldots.}$
$A=\mathbb{N}-{1,2,3,4,6}$
$\therefore P(B)=1-[P(w_1)+P(w_2)+P(w_3)+P(w_4)+P(w_6)]$
$=1-[\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{64}]$
$=1-\frac{32+16+8+4+1}{64}=\frac{3}{64}$
