Five students are to be selected from a large population. If 60% of them are boys and the rest are girls, find the probability that :
(a) exactly 3 of them are boys;
(b) at least 3 of them are girls.
60% boys, 40% girls.
p(boy) = p = 60% = 0.6
p(girl) = q = 40% = 0.4
The binomial distribution function is
\((p + q)^{5} = p^{5} + 5p^{4} q + 10p^{3} q^{2} + 10p^{2} q^{3} + 5pq^{4} + q^{5}\)
(a) p(exactly 3 boys) = \(10p^{3} q^{2}\)
\(10(\frac{3}{5})^{3} (\frac{2}{5})^{2}\)
= \(10(\frac{27}{125})(\frac{4}{25})\)
= \(\frac{1080}{3125}\)
= \(\frac{216}{625}\)
(b) p(at least 3 are girls) = 1 - [p(0 girls) + p(1 girl) + p(2 girls)].
= \(1 - [p^{5} + 5p^{4}q + 10p^{3} q^{2}]\)
= \(1 - [(\frac{3}{5})^{5} + 5(\frac{3}{5})^{4} (\frac{2}{5}) + 10(\frac{3}{5})^{3} (\frac{2}{5})^{2}]\)
= \(1 - [\frac{243}{3125} + \frac{810}{3125} + \frac{1080}{3125}]\)
= \(1 - [\frac{2133}{3125}]\)
= \(\frac{992}{3125}\).
Contributions ({{ comment_count }})
Please wait...
Modal title
Report
Block User
{{ feedback_modal_data.title }}