Question

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

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

The probability of getting a head in a single toss of a coin is $p=\frac{1}{2}$(say). The probability of getting 5 or 6 in a single throw of a die is $q=\frac{2}{6}=\frac{1}{3}$(say). Therefore, the required probability is
$p+(1-p)(1-q)p+(1-p)(1-q)(1-p)(1-q)p+....$
$=p+(1-p)(1-q)p+(1-p{)}^2(1-q{)}^{2}p+....$
$=\frac{1}{1-(1-p)(1-q)}$
$=\frac{\frac{1}{2}}{1-\frac{1}{2}\times \frac{2}{3}}=\frac{3}{4}$