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Q.
From a group of $5$ boys and $3$ girls three persons are chosen at random; Find the probability that there are more girls than boys
Probability - Part 2
Solution:
Three persons can be chosen out of 8 in $^8C_3 = 56$ ways. The number of girls is more than that of the boys if either 3 girls are chosen or two girls and one boy is chosen. This can be
done in $^{3}C_{3} + ^{3}C_{2} \times^{5}C_{1}$
$ = 1 + 3 \times 5 = 16$ ways.
$\therefore $ Required probability $= \frac{16}{56}= \frac{2}{7}$