Question

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

• A. $\frac{3}{4}$
• B. $\frac{1}{2}$
• C. $\frac{1}{3}$
• D. $None$

Answer 1

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

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