The probability of a boy winning a game is $\frac{2}{3}$. Let n denotes that the least number of times he must play the game so that the probability of winning the game at least once is more than 90% and X denotes that number of times he win the game. Hence n, mean, variance and standard deviation of random variable X are respectively:
(A) 3
(B) $\frac{2}{3}$
(C) 2
(D) 0.81
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A), (C), (B), (D)
$\text{P(not winning in one attempt)} = 1-\frac{2}{3}=\frac{1}{3}$
$\text{P(not winning in all n games)} = \left(frac{1}{3}\right)^n$
$\text{P(atleast winning once) = 1 - P(not winning in all n games)}$
$=1-\left(frac{1}{3}\right)^n$
$1-\left(frac{1}{3}\right)^n>0.9$ [given]
$⇒\left(frac{1}{3}\right)^n<0.1$
By trial & error method
$n=3$ $(∵\left(frac{1}{3}\right)^3=\frac{1}{27}<0.1)$
Mean of a binomial random variable
$μ=n.p$
$=3.\frac{2}{3}=2$
Variance = $n.p.q$
$=3.\frac{2}{3}.\frac{1}{3}$
$=\frac{2}{3}$
Standard deviation, $\sqrt{Var(x)}=\sqrt{\frac{2}{3}}$
$n=3$, Mean = 2, Variance = $\frac{2}{3}$, Standard deviation = $\sqrt{\frac{2}{3}}$