**Question **: A man alternately tosses a coin and throw a die beginning with the coin. The probability that he gets a head in the coin before he gets a 5 or 6 in the dice is

(A) 3/4

(B) 1/2

(C) 1/3

(D) none of these

**Answer **: option (A)

**Solution **:

The probability of getting a head in a single toss of a coin is p=1/2 (say).

The probability of getting 5 or 6 in a single throw of a die is q=2/6 =1/3 (say).

Therefore, the required probability is

p + (1-p) (1-q) p + (1-p) (1-q) (1-p) (1-q) p + …..

\begin{aligned}\Rightarrow p+\left( 1-p\right) \left( 1-q\right) p+\left( 1-p\right) ^{2}\left( 1-q\right) ^{2}p+\ldots \ldots \ \Rightarrow \dfrac{p}{1-\left( 1-p\right) \left( 1-q\right) }=\dfrac{\dfrac{1}{2}}{1-\dfrac{1}{2}\times \dfrac{2}{3}}=\dfrac{3}{4}\end{aligned}