There are 6 boys and 8 girls in a class. If five students are selected from the class, find the probability that more girls than boys are selected
Total number of students =6 boys + 8 girls = 14 students.
Total ways to choose 5 students out of 14:
Total ways = \(^{14}C_5=\frac{14!}{5!*(14 - 5)!}=2002.\)
The number of ways to have more girls than boys:
(Selecting 3 girls and 2 boys) or (Selecting 4 girls and 1 boy)
Selecting 3 girls and 2 boys:\(^8C_3\times ^6C_2\)
=(\(\frac{8!}{3!*(8 - 3)!})\times(\frac{6!}{2! * (6 - 2)!})=56\times 15=840.\)
Selecting 4 girls and 1 boy:\(^8C_4\times^6C_1\)
\(=(\frac{8!}{4!*(8 - 4)!})\times(\frac{6!}{1! * (6 - 1)!})=70\times 6=420.\)
Total favorable cases =840+420=1260.
Finally, the probability is given by:
Probability (more girls than boys)
= \(\frac{Total Favorable Cases}{Total Ways}=\frac{1260}{2002}=\frac{90}{143}≈0.629.\)
The probability that more girls than boys are selected is approximately 0.629, or about 62.9%.
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