Find the probability distribution of X, the number of heads in three…

The correct answer is Option (1) → 

$X$	0	1	2	3
$P(X)$	$\frac{1}{8}$	$\frac{3}{8}$	$\frac{3}{8}$	$\frac{1}{8}$

The sample space associated with this experiment is

$S = \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\}$

All 8 outcomes are equally likely.

Let random variable X be defined as the number of heads, then X can take the 0, 1, 2 and 3.

$\text{P(X=0)=P(TTT)}=\frac{1}{8}$

$\text{P(X=1)=P(TTH or THT or HTT)}=\frac{3}{8}$

$\text{P(X=2)=P(HHT or HTH or THH)}=\frac{3}{8}$

$\text{P(X=3)=P(HHH)}=\frac{1}{8}$

Hence, the required probability distribution is

$X$	0	1	2	3
$P(X)$	$\frac{1}{8}$	$\frac{3}{8}$	$\frac{3}{8}$	$\frac{1}{8}$