A family consists of 8 persons. If 4 persons | vedclass

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(D) Let the number of men be $m$ and the number of women be $w$. Given $m + w = 8$. The probability of choosing $2$ men and $2$ women from the family

Vedclass · Accepted Answer

$\frac{2}{7}$

Vedclass · Answer

(D) Let the number of men be $m$ and the number of women be $w$. Given $m + w = 8$. The probability of choosing $2$ men and $2$ women from the family is given by $P(E) = \frac{\binom{m}{2} \binom{w}{2}}{\binom{8}{4}}$. We are given that this event has occurred. We want to find the probability that $m = w = 4$. If $m = 4$ and $w = 4$,the probability of picking $2$ men and $2$ women is $P(E|m=4, w=4) = \frac{\binom{4}{2} \binom{4}{2}}{\binom{8}{4}} = \frac{6 	imes 6}{70} = \frac{36}{70}$. Assuming all possible compositions $(m, w)$ are equally likely,where $m+w=8$,the possible pairs $(m, w)$ are $(0,8), (1,7), (2,6), (3,5), (4,4), (5,3), (6,2), (7,1), (8,0)$. Using Bayes' Theorem,the probability that $m=4, w=4$ given the observation is $\frac{P(E|m=4, w=4)}{\sum P(E|m_i, w_i)} = \frac{36}{\sum \binom{m_i}{2} \binom{w_i}{2}}$. Calculating the sum: $\binom{0}{2}\binom{8}{2} + \binom{1}{2}\binom{7}{2} + \binom{2}{2}\binom{6}{2} + \binom{3}{2}\binom{5}{2} + \binom{4}{2}\binom{4}{2} + \binom{5}{2}\binom{3}{2} + \binom{6}{2}\binom{2}{2} + \binom{7}{2}\binom{1}{2} + \binom{8}{2}\binom{0}{2} = 0 + 0 + 15 + 30 + 36 + 30 + 15 + 0 + 0 = 126$. The required probability is $\frac{36}{126} = \frac{2}{7}$.

Box	Red	White	Black
$A$	$1$	$6$	$3$
$B$	$6$	$2$	$2$
$C$	$8$	$1$	$1$
$D$	$0$	$6$	$4$

$A$ family consists of $8$ persons. If $4$ persons are chosen at random and they are found to be $2$ men and $2$ women,then the probability that there are equal number of men and women in that family is

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