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 ?

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) (1/4) (B) (1/2) (C) (7/12) (D) (8/15)

Tardigrade Admin · Accepted Answer

P(A win )=((1/2))+((1/2))4((1/2))+((1/2))8((1/2))+ ldots =((1/2)/1-((1)2)4)=((1/2)/(15)16)=(8/15)

Q. 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 ?

$\frac{1}{4}$

$\frac{1}{2}$

$\frac{7}{12}$

$\frac{8}{15}$

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

Q. 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 ?

41​

21​

127​

158​