Question

A dice is thrown (2n + 1) times. The probability of getting 1, 3 or 4 at most n times, is

• A. $\frac{1}{2}$
• B. $\frac{1}{3}$
• C. $\frac{1}{4}$
• D. None of these

Answer 1

The correct option is A $\frac{1}{2}$

Let X be the number of times 1, 3 or 4 occur on the die. Then X follows a binomial distribution with parameter and $p=\frac{3}{6}=\frac{1}{2}$
We have P(1, 3 or 4 occur at most n times on the die)
$=P(0\le X\le n)=P(X=0)+P(X=1)+.....+P(X=n)$
$={}^{2n+1}{C}_0(\frac{1}{2}{)}^{2n+1}+{}^{2n+1}{C}_1(\frac{1}{2}{)}^{2n+1}+......+{}^{2n+1}{C}_n(\frac{1}{2}{)}^{2n+1}$
$=[{}^{2n+1}{C}_0+{}^{2n+1}{C}_1+....+{}^{2n+1}{C}_n](\frac{1}{2}{)}^{2n+1}$
Let $S={}^{2n+1}{C}_0+{}^{2n+1}{C}_1+....+{}^{2n+1}{C}_n$
$\Rightarrow 2S=2.{}^{2n+1}{C}_0+2.{}^{2n+1}{C}_1+.....+2.{}^{2n+1}{C}_n$
$=({}^{2n+1}{C}_0+{}^{2n+1}{C}_{2n+1})+({}^{2n+1}{C}_1+{}^{2n+1}{C}_{2n})+......+({}^{2n+1}{C}_n+{}^{2n+1}{C}_{n+1})$
$\Rightarrow S=2^{2n}$
Hence required probability = $2^{2n}(\frac{1}{2}{)}^{2n+1}=\frac{1}{2}$