Four persons A, B, C and D throw an unbiased | vedclass

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(D) The probability of getting an even number on a die is $P(E) = \frac{3}{6} = \frac{1}{2}$,and the probability of getting an odd number is $P(O) = \

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$\frac{8}{15}$

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(D) The probability of getting an even number on a die is $P(E) = \frac{3}{6} = \frac{1}{2}$,and the probability of getting an odd number is $P(O) = \frac{3}{6} = \frac{1}{2}$.$A$ wins if $A$ gets an even number on the $1^{st}, 5^{th}, 9^{th}, \dots$ turn.Let $p = \frac{1}{2}$ be the probability of success and $q = \frac{1}{2}$ be the probability of failure.$A$ wins on the $1^{st}$ turn with probability $p = \frac{1}{2}$.$A$ wins on the $5^{th}$ turn if $A, B, C, D$ fail on their first turns and $A$ succeeds on the $5^{th}$ turn: $q^4 p = (\frac{1}{2})^4 	imes \frac{1}{2} = (\frac{1}{2})^5$.$A$ wins on the $9^{th}$ turn with probability $q^8 p = (\frac{1}{2})^8 	imes \frac{1}{2} = (\frac{1}{2})^9$.This is an infinite geometric series with first term $a = \frac{1}{2}$ and common ratio $r = q^4 = (\frac{1}{2})^4 = \frac{1}{16}$.The sum of the series is $S = \frac{a}{1-r} = \frac{1/2}{1 - 1/16} = \frac{1/2}{15/16} = \frac{1}{2} 	imes \frac{16}{15} = \frac{8}{15}$.

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

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