There are 8 boys and 7 girls in a classroom. | vedclass

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(C) Total number of children = $8 + 7 = 15$. Total ways to choose $3$ slips from $15$ is $^{15}C_3 = \frac{15 	imes 14 	imes 13}{3 	imes 2 	imes 1

Vedclass · Accepted Answer

$\frac{4}{5}$

Vedclass · Answer

(C) Total number of children = $8 + 7 = 15$. Total ways to choose $3$ slips from $15$ is $^{15}C_3 = \frac{15 	imes 14 	imes 13}{3 	imes 2 	imes 1} = 455$. Case $1$: One boy and two girls. Ways = $^8C_1 	imes ^7C_2 = 8 	imes \frac{7 	imes 6}{2} = 8 	imes 21 = 168$. Case $2$: One girl and two boys. Ways = $^7C_1 	imes ^8C_2 = 7 	imes \frac{8 	imes 7}{2} = 7 	imes 28 = 196$. Total favorable ways = $168 + 196 = 364$. Probability = $\frac{364}{455}$. Dividing both by $91$,we get $\frac{364 \div 91}{455 \div 91} = \frac{4}{5}$.

There are $8$ boys and $7$ girls in a classroom. If the names of all those children are written on paper slips and $3$ slips are drawn at random from them,then the probability of getting the names of one boy and two girls or one girl and two boys is

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