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

\(\begin{array}{l}H H H-3 H \text { and } 0 T \\ H H T-2 H \text { and } 1 T=|n(H)-n(T)|=1 \\ H T H-2 H \text { and } 1 T=|n(H)-n(T)|=1 \\ H T T-1 H \text { and } 2 T=|n(H)-n(T)|=1 \\ T H H-2 H \text { and } 1 T=|n(H)-n(T)|=1 \\ T H T-2 T \text { and } 1 H=|n(H)-n(T)|=1 \\ T T H-2 T \text { and } 1 H=|n(H)-n(T)|=1 \\ T T T-3 T \\ P(X=1)=\frac{6}{8}=\frac{3}{4}\end{array}\)
Or (second method)
Probability distribution of X
\(\begin{array}{|l|l|l|l|l|} \hline \text { Out come } & 3 \text{T} & 2 \text{T} 1 \text{H} & 1 \text{T} 2 \text{H} & 3 \text{H} \\ \hline \text{X} & 3 & 1 & 1 & 3 \\ \hline \text{P} & \frac{1}{8} & \frac{3}{8} & \frac{3}{8} & \frac{1}{8} \\ \hline \end{array}\)
\(\Rightarrow P ( X =1)=\frac{3}{8}+\frac{3}{8}=\frac{3}{4}\)