Question

A man alternately tosses a coin and throws a die beginning with coin. The probability that he gets a head in the coin before he gets 5 or 6 on the die is

• A. $\frac{3}{4}$
• B. $\frac{1}{2}$
• C. $\frac{1}{3}$
• D. $\frac{1}{4}$

Answer 1

The correct option is A $\frac{3}{4}$

probability of getting head $=\frac{1}{2}$ and probability of throwing 5or 6 with a die $=\frac{2}{6}=\frac{1}{3}.$
He starts with coin and alternatively tosses the coin and throw the die and he win if he get a
head before he get 5 or 6
Hence required probability $=\frac{1}{2}+(\frac{1}{2}.\frac{2}{3})\frac{1}{2}+(\frac{1}{2}.\frac{2}{3})(\frac{1}{2}.\frac{2}{3})\frac{1}{2}+...$
$=\frac{1}{2}[1+\frac{1}{3}+{\left(\frac{1}{3}\right)}^2+...]$
$=\frac{1}{2}.\left(\frac{1}{1-\frac{1}{3}}\right)=\frac{3}{4}.$