An unfair coin is two times as likely to show a head than a tail. The coin is tossed once. If head comes it is tossed once more, but if tail comes, then a fair die is thrown. What is the probability of getting a prime digit on the die, given that there is at least one tail?

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$\frac{3}{5}$

$\frac{3}{10}$

$\frac{2}{5}$

$\frac{1}{3}$

answer is B.

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Detailed Solution

Req. probability $= \frac{P (T 2) + P (T 3) + P (T 5)}{1 - P (H H)} = \frac{3 \times \frac{1}{3} \times \frac{1}{6}}{1 - \frac{2}{3} \times \frac{2}{3}} = \frac{3}{10}$

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