Question

A coin is tossed three times. If X denotes the absolute difference between the number of heads and the number of tails then P(X = 1) =

• A. $\frac{1}{2}$
• B. $\frac{2}{3}$
• C. $\frac{1}{6}$
• D. $\frac{3}{4}$

Answer 1

Answer: (D)

$\frac{3}{4}$

Answer 2

Given, A coin is tossed three times
and $X=$ absolute difference between the number of heads and number of tails.
Now, $X=1$ when exactly two head or two tail comes
$\therefore P(X=1)=\frac{3}{8}+\frac{3}{8}=\frac{6}{8}=\frac{3}{4}$
$[\because$ probability of exacty two head $=\frac{3}{8}$,
and probability of exactly two tails $=\frac{3}{8} ]$