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Question

(C) $P(W) = \frac{1}{3}, P(L) = \frac{2}{3}$. Let $x$ be the number of wins and $y$ be the number of losses. Given $x+y=10$ and $|x-y| \leq 2$.Case $I

Vedclass · Accepted Answer

$8288$

Vedclass · Answer

(C) $P(W) = \frac{1}{3}, P(L) = \frac{2}{3}$. Let $x$ be the number of wins and $y$ be the number of losses. Given $x+y=10$ and $|x-y| \leq 2$.Case $I$: $|x-y|=0 \Rightarrow x=y$. Since $x+y=10$,we have $x=5, y=5$. The probability is $P(x=5) = {}^{10}C_5 (\frac{1}{3})^5 (\frac{2}{3})^5 = {}^{10}C_5 \frac{2^5}{3^{10}}$.Case $II$: $|x-y|=1$. Since $x+y=10$,$x-y = \pm 1$ implies $2x = 11$ or $2x = 9$,which has no integer solutions. Thus,$P(|x-y|=1) = 0$.Case $III$: $|x-y|=2$. This implies $x-y=2$ or $x-y=-2$. If $x-y=2$ and $x+y=10$,then $x=6, y=4$. If $x-y=-2$ and $x+y=10$,then $x=4, y=6$.$P(|x-y|=2) = P(x=6) + P(x=4) = {}^{10}C_6 (\frac{1}{3})^6 (\frac{2}{3})^4 + {}^{10}C_4 (\frac{1}{3})^4 (\frac{2}{3})^6 = {}^{10}C_6 \frac{2^4}{3^{10}} + {}^{10}C_4 \frac{2^6}{3^{10}}$.Total probability $p = P(|x-y|=0) + P(|x-y|=2) = \frac{{}^{10}C_5 2^5 + {}^{10}C_6 2^4 + {}^{10}C_4 2^6}{3^{10}}$.$3^9 p = \frac{{}^{10}C_5 2^5 + {}^{10}C_6 2^4 + {}^{10}C_4 2^6}{3} = \frac{252 	imes 32 + 210 	imes 16 + 210 	imes 64}{3} = \frac{8064 + 3360 + 13440}{3} = \frac{24864}{3} = 8288$.

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