If a man throws a die until he gets a number | vedclass

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(A) Let $E$ be the event of getting a number greater than $3$ in a single throw of a die. The outcomes are $\{4, 5, 6\}$.$P(E) = \frac{3}{6} = \frac{1

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$1/3$

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(A) Let $E$ be the event of getting a number greater than $3$ in a single throw of a die. The outcomes are $\{4, 5, 6\}$.$P(E) = \frac{3}{6} = \frac{1}{2}$.Let $F$ be the event of getting a $5$ in a single throw. $P(F) = \frac{1}{6}$.Let $S$ be the event of getting a number $\leq 3$ in a single throw. $P(S) = \frac{3}{6} = \frac{1}{2}$.The man stops when he gets a number $> 3$. The possible sequences for the last throw being $5$ are:$1$. First throw is $5$: Probability $= \frac{1}{6}$.$2$. First throw is $\leq 3$,second throw is $5$: Probability $= \frac{1}{2} 	imes \frac{1}{6}$.$3$. First two throws are $\leq 3$,third throw is $5$: Probability $= (\frac{1}{2})^2 	imes \frac{1}{6}$.This is an infinite geometric series: $\frac{1}{6} + \frac{1}{6}(\frac{1}{2}) + \frac{1}{6}(\frac{1}{2})^2 + \dots$The sum is $S = \frac{a}{1-r} = \frac{1/6}{1-1/2} = \frac{1/6}{1/2} = \frac{1}{3}$.

If a man throws a die until he gets a number bigger than $3$,then the probability that he gets a $5$ in his last throw is

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