Question 2

Shanu and Ishu play a tossing game. They alternatively toss a coin. Whoever receives 'Head' first, wins the game. What is Ishu's chance of winning the game if Shanu gets a chance to toss coin first?

Solution

Probability of Ishu winning in first round = P(Shanu tail)P(Ishu head) = $$\frac{1}{2}\cdot\frac{1}{2}$$ = $$\left(\frac{1}{2}\right)^2$$

Probability of Ishu winning in second round = P(Shanu tail)P(Ishu tail)P(Shanu tail)P(Ishu head) = $$\left(\frac{1}{2}\right)^4$$

Similarly, calculating for further rounds we get

Probability = $$\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^4+\left(\frac{1}{2}\right)^{6\ }+\ ......\infty\$$

Sum of infinite terms in GP(first term $$a$$ and common difference $$r$$) = $$\frac{a}{1-r}$$

So, in the above sum, $$a=\left(\frac{1}{2}\right)^{^2},\ r\ =\ \left(\frac{1}{2}\right)^2$$

So, Probability = $$\frac{(1/4)}{(1-1/4)}=\frac{1}{3}$$

Thus, the correct option is A.