Question

Four persons $A, B, C$ and $D$ throw an unbiased die, turn by turn, in succession till one gets an even number 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

Answer: (D)

$P(A$ win $)=\left(\frac{1}{2}\right)+\left(\frac{1}{2}\right)^{4}\left(\frac{1}{2}\right)+\left(\frac{1}{2}\right)^{8}\left(\frac{1}{2}\right)+\ldots$
$=\frac{\frac{1}{2}}{1-\left(\frac{1}{2}\right)^{4}}=\frac{\frac{1}{2}}{\frac{15}{16}}=\frac{8}{15}$