In a tournament,there are 12 players P_1, P_2 | vedclass

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Question

(B) There are $12$ players to be divided into $6$ pairs. The total number of ways to pair $12$ players is $\frac{12!}{2^6 	imes 6!}$.Case $1$: $P_1$

Vedclass · Accepted Answer

$\frac{6}{11}$

Vedclass · Answer

(B) There are $12$ players to be divided into $6$ pairs. The total number of ways to pair $12$ players is $\frac{12!}{2^6 	imes 6!}$.Case $1$: $P_1$ and $P_2$ are in the same pair.The probability that $P_1$ and $P_2$ are paired together is $\frac{1}{11}$.If they are in the same pair,one must win and one must lose. Thus,the probability that exactly one of them loses is $1 	imes \frac{1}{11} = \frac{1}{11}$.Case $2$: $P_1$ and $P_2$ are in different pairs.The probability that $P_1$ and $P_2$ are not paired together is $1 - \frac{1}{11} = \frac{10}{11}$.Let $P_1$ be in pair $A$ and $P_2$ be in pair $B$. For exactly one of them to lose,either ($P_1$ loses and $P_2$ wins) or ($P_1$ wins and $P_2$ loses).The probability of each outcome in a game is $\frac{1}{2}$.Probability $= \frac{10}{11} 	imes (\frac{1}{2} 	imes \frac{1}{2} + \frac{1}{2} 	imes \frac{1}{2}) = \frac{10}{11} 	imes \frac{1}{2} = \frac{5}{11}$.Total probability $= \frac{1}{11} + \frac{5}{11} = \frac{6}{11}$.

List-$I$	List-$II$
$A. P(E_2) =$	$I. 2/3$
$B. P(E_1 \| E_2) =$	$II. 5/6$
$C. P(\bar{E}_2 \| E_1) =$	$III. 1/3$
$D. P(\bar{E}_1 \cup \bar{E}_2) =$	$IV. 1/2$

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