Question

A fair coin is tossed 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. 1/64
• B. 1/32
• C. 1/16
• D. 1/8

Answer 1

Answer: (B)

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,
$P(X=3)=P(X=5)$
$\Rightarrow {}^n{C}_3{\left(\frac{1}{2}\right)}^3{\left(\frac{1}{2}\right)}^{n-3}={}^n{C}_5{\left(\frac{1}{2}\right)}^5{\left(\frac{1}{2}\right)}^{n-5}$
$\Rightarrow {}^n{C}_3={}^n{C}_6$
$\Rightarrow n=3+5\left[\because {}^n{C}_x={}^n{C}_y\Rightarrow x+y=n\right]$
$\Rightarrow n=8$
Now, $P(X=1)={}^8{C}_1{\left(\frac{1}{2}\right)}^1{\left(\frac{1}{2}\right)}^{8-1}$
$={}^8{C}_1\times {\left(\frac{1}{2}\right)}^8=\frac{1}{32}$