Question

A coin is tossed $n$ times, the probability that head will turn up an even number of times is

• A. $\frac{n+1}{2n}$
• B. $\frac{n}{n+1}$
• C. $\frac{1}{2}$
• D. $2^{n-1}$

Answer 1

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

Let x denote the number of heads in n trials then $P(x=r){=}^n{C}_r{p}^r{q}^{n-r}{=}^n{C}_r\left(\frac{1}{2}{)}^r\right(\frac{1}{2}{)}^{n-r}{=}^n{C}_r(\frac{1}{2}{)}^n$

now required probability is $p(x=0,2,4,6,\cdots )\therefore P(X=0,2,4,6,\cdots )=P\left(0\right)+P\left(2\right)+P\left(4\right)+\cdots =\left(\frac{1}{2}{)}^n\right[c_0+c_2+c_4+\cdots ]=(\frac{1}{2}{)}^{2n-1}=\frac{1}{2}$