Question

A man alternatively tosses a coin and throws a die. The probability of getting a head on the coin before he gets 4 on the die is

• A. $\frac{6}{7}$
• B. $\frac{2}{3}$
• C. $\frac{3}{4}$
• D. $\frac{1}{2}$

Answer 1

The correct option is A $\frac{6}{7}$

Let the desired event be called E
then $P\left(E\right)=P\left(H\right)+P\left(T\right)P\left(\overline{4}\right)P\left(H\right)+P\left(T\right)P\left(\overline{4}\right)P\left(T\right)P\left(\overline{4}\right)P\left(H\right)+.............\infty$
where $P\left(\overline{4}\right)=\frac{5}{6},P\left(H\right)=P\left(T\right)=0.5$
thus we can see that an G.P sum is formed with first term =P(H) and common ratio =$P\left(T\right)P\left(\overline{4}\right)$
hence $P\left(E\right)=\frac{P\left(H\right)}{1-P\left(T\right)P\left(\overline{4}\right)}=\frac{0.5\times 12}{7}=\frac{6}{7}$