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Probability

Section: Mathematics

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Question : 3 of 92

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There are 8 boys and 7 girls in a class room. 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

[AP EAPCET 23 May 2025 S2]

$\;\frac{1}{5}$
$\;\frac{3}{4}$
$\;\frac{4}{5}$
$\;\frac{1}{4}$

Solution:

Total childrens

=15(8 B+7 G)

Total ways of choosing 3 childrens out of 15 is

={}^{15} C_3=\;\frac{15 \times 14 \times 13}{3 \times 2 \times 1}=455

Case I Choosing 1 boy and 2 girls

={}^8 C_1 \cdot {}^7 C_2=\;\frac{8 \times 7 \times 6}{2}=168

Case II Choosing 2 boys and 1 girl

={}^8 C_2 \cdot {}^7 C_1=\;\frac{8 \times 7}{2} \times 7=196

Thus, required probability

=\;\frac{168+196}{455}=\;\frac{364}{455}=\;\frac{4}{5}

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