A coin is tossed three times. If $\mathrm{X}$ denote the absolute difference between the number of heads and…
A coin is tossed three times. If $\mathrm{X}$ denote the absolute difference between the number of heads and the number of tails, then, $\mathrm{P}(\mathrm{X}=1)=$
$\frac{1}{6}$
$\frac{1}{2}$
$\frac{2}{3}$
$\frac{3}{4}$
Solution
A coin is tossed 3 times
$\Rightarrow \mathrm{n}(\mathrm{S})=8$
Possibilities are : $(1 \mathrm{H}, 2 \mathrm{~T}),(2 \mathrm{H}, 1 \mathrm{~T}),(3 \mathrm{H}, 0 \mathrm{~T}),(0 \mathrm{H}, 3 \mathrm{~T})$
Thus values of $\mathrm{X}$ can be 1 and 3 .
Now $(1 \mathrm{H}, 2 \mathrm{~T})$ can occur in 3 ways. Also $(2 \mathrm{H}, 1 \mathrm{~T})$ can occur in 3 ways.
$\therefore \mathrm{P}(\mathrm{X}=1)=\frac{6}{8}=\frac{3}{4}$