Two persons A and B are throwing an unbiased | vedclass

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Question

(A) The probability of getting a $3$ in a single throw is $p = \frac{1}{6}$.The probability of not getting a $3$ is $q = 1 - \frac{1}{6} = \frac{5}{6}

Vedclass · Accepted Answer

$\frac{6}{11}, \frac{5}{11}$

Vedclass · Answer

(A) The probability of getting a $3$ in a single throw is $p = \frac{1}{6}$.The probability of not getting a $3$ is $q = 1 - \frac{1}{6} = \frac{5}{6}$.$A$ starts the game. $A$ wins if $A$ gets a $3$ on the $1^{st}, 3^{rd}, 5^{th}, \dots$ throw.$P(A) = p + q^2p + q^4p + \dots = \frac{p}{1 - q^2}$.Substituting the values: $P(A) = \frac{1/6}{1 - (5/6)^2} = \frac{1/6}{1 - 25/36} = \frac{1/6}{11/36} = \frac{6}{11}$.Since the total probability is $1$,$P(B) = 1 - P(A) = 1 - \frac{6}{11} = \frac{5}{11}$.Thus,the probabilities are $\frac{6}{11}$ and $\frac{5}{11}$.

Two persons $A$ and $B$ are throwing an unbiased six-faced dice alternatively,with the condition that the person who throws $3$ first wins the game. If $A$ starts the game,then the probabilities of $A$ and $B$ winning the game are,respectively:

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