Question

A fair coin is tossed at a fixed number of times. If the probability of getting exactly $3$ heads equals the probability of getting exactly $5$ heads, then the probability of getting exactly one head is

• A. $\frac{1}{64}$
• B. $\frac{1}{32}$
• C. $\frac{1}{16}$
• D. $\frac{1}{8}$

Answer 1

The correct option is B $\frac{1}{32}$

Let the coin be tossed $n$ times.

Let getting head is consider to be success.
$\therefore p=\frac{1}{2},q=1-p=1-\frac{1}{2}=\frac{1}{2}$
It is given that, $p(X=3)=p(x=5)$

$\Rightarrow {}^n{C}_3{\left(\frac{1}{2}\right)}^3{\left(\frac{1}{2}\right)}^{n-3}={}^n{C}_5{\left(\frac{1}{2}\right)}^5{\left(\frac{1}{2}\right)}^{n-5}$
${\Rightarrow }^n{C}_3{=}^n{C}_5$

$\Rightarrow n=3+5$ $[{\because }^n{C}_x{=}^n{C}_y\Rightarrow x+y=n]$
$\Rightarrow n=8$

Now, $P(x=1)={}^8{C}_1{\left(\frac{1}{2}\right)}^1{\left(\frac{1}{2}\right)}^{8-1}$

${=}^8{C}_1\times {\left(\frac{1}{2}\right)}^8=\frac{1}{32}$