Question

Four persons $A,B,C$ and $D$ throw an unbiased die, turn by turn, in succession till one gets an even numbers and win the game. What is the probability that $A$ wins if $A$ begins?

• A. $\frac{1}{4}$
• B. $\frac{1}{2}$
• C. $\frac{7}{12}$
• D. $\frac{8}{15}$

Answer 1

The correct option is D $\frac{8}{15}$

$A$ wins in $1^{st}$ attempt $P\left(\text{even number}\right)=\frac{1}{2}$, $P\left(\text{odd number}\right)=\frac{1}{2}$

$\Rightarrow P\left(A\right)+P\left(\overline{A}\right)P\left(\overline{B}\right)P\left(\overline{C}\right)P\left(\overline{D}\right)P\left(A\right)+....\Rightarrow \frac{1}{2}+\frac{1}{2}\cdot \frac{1}{2}\cdot \frac{1}{2}\cdot \frac{1}{2}\cdot \frac{1}{2}+\cdots \Rightarrow \left(\frac{1}{2}\right)+{\left(\frac{1}{2}\right)}^5+{\left(\frac{1}{2}\right)}^9+\cdots ,S_{\infty }=\frac{\frac{1}{2}}{1-{\left(\frac{1}{2}\right)}^4}=\frac{\frac{16}{2}}{16-1}=\frac{8}{15}$