A coin is biased so that the head is 3 times as likely to occur as...

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

$∴ P ( H ) = \dfrac{3}{4}, P ( T ) = \dfrac{1}{4}$

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

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

Now we know that,

Mean is given by $\mathop{∑}_{i = 1}^{n}{X}_{i}$P\left({X}_{i}\right)$</p><p>Mean$= 1 \times \dfrac{3}{4} + 2 \times \dfrac{3}{16} + 3 \times \dfrac{1}{16}$$= \dfrac{21}{16}

Hence, this is the correct option.