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(C) Let $W$ denote winning (getting a number $> 4$,probability $p = \frac{2}{6} = \frac{1}{3}$) and $L$ denote losing (getting a number $\le 4$,probab

Vedclass · Accepted Answer

$\frac{19}{9}$

Vedclass · Answer

(C) Let $W$ denote winning (getting a number $> 4$,probability $p = \frac{2}{6} = \frac{1}{3}$) and $L$ denote losing (getting a number $\le 4$,probability $q = \frac{4}{6} = \frac{2}{3}$). The player quits when $W$ occurs or after $3$ throws. The possible outcomes are:$(i)$ $W$: Win $5$ rupees. Probability $P_1 = \frac{1}{3}$.(ii) $LW$: Lose $1$ rupee,then win $5$ rupees. Net gain $= 4$ rupees. Probability $P_2 = \frac{2}{3} 	imes \frac{1}{3} = \frac{2}{9}$.(iii) $LLW$: Lose $1$ rupee,lose $1$ rupee,then win $5$ rupees. Net gain $= 3$ rupees. Probability $P_3 = \frac{2}{3} 	imes \frac{2}{3} 	imes \frac{1}{3} = \frac{4}{27}$.(iv) $LLL$: Lose $1$ rupee,lose $1$ rupee,lose $1$ rupee. Net gain $= -3$ rupees. Probability $P_4 = \frac{2}{3} 	imes \frac{2}{3} 	imes \frac{2}{3} = \frac{8}{27}$.The expected value $E = \sum x_i P_i = 5(\frac{1}{3}) + 4(\frac{2}{9}) + 3(\frac{4}{27}) + (-3)(\frac{8}{27})$.$E = \frac{5}{3} + \frac{8}{9} + \frac{12}{27} - \frac{24}{27} = \frac{45 + 24 + 12 - 24}{27} = \frac{57}{27} = \frac{19}{9}$.

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