Question

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

• A.
• B.
• C.
• D. None of these

Answer 1

The correct option is A

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)
$=P(0\le X\le n)=P(X=0)+P(X=1)+\dots \dots +P(X=n)$
${=}^{2n+1}{C}_0{\left(\frac{1}{2}\right)}^{2n+1}{+}^{2n+1}{C}_1{\left(\frac{1}{2}\right)}^{2n+1}+....{+}^{2n+1}{C}_n{\left(\frac{1}{2}\right)}^{2n+1}$
$[{=}^{2n+1}{C}_0{+}^{2n+1}{C}_1+....{+}^{2n+1}{C}_n]{\left(\frac{1}{2}\right)}^{2n+1}$
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 = $2^{2n}{\left(\frac{1}{2}\right)}^{2n+1}=\frac{1}{2}$