A coin is biased so that the head is 3 times as likely to occur as tail. This coin is tossed until a head or…

It is given that $P\left(H\right)=3P\left(T\right)$

$\therefore P\left(H\right)=\frac{3}{4},P\left(T\right)=\frac{1}{4}$

Since coin is tossed till either $1H$ or $3T$ occurs. So, process will end in the maximum of $3$ throws.

$Xi$	$1$	$2$	$3$
$P\left(Xi\right)$	$\frac{3}{4}$	$\frac{1}{4}·\frac{3}{4}$	$\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}+\frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}$

Now we know that,

Mean is given by $\sum _{i=1}^n{X}_{i}P\left(X_i\right)$ 

Mean$=1\times \frac{3}{4}+2\times \frac{3}{16}+3\times \frac{1}{16}$$=\frac{21}{16}$

Hence, this is the correct option.