Question

A die is thrown $2n+1$ times. The probability of getting $1$ or $3$ or $4$ atmost $n$ times is

• A. $\frac{1}{2}$
• B. $\frac{1}{n}$
• C. $\frac{n}{(2n+1)}$
• D. $\frac{1}{(2n+1)}$

Answer 1

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

Let $P$ be the probability of getting $1/3/4$ in a throw

$q=1-P$

Using binomial theorem,

Required probability${=}^{2n+1}{C}_0{P}^0{q}^{2n+1}$${+}^{2n+1}{C}_1{P}^1{q}^{2n+1-1}$$+........{+}^{2n+1}{C}_n{P}^n{q}^{2n+1-n}$

Here we have $P=q=0.5$

Hence by solving above equation, we get

Net probability$=\frac{1}{2}$