p(have valid license) = p = 0.8
p(no valid license) = q = 0.2.
The binomial probability distribution is
\((p + q)^{6} = p^{6} + 6p^{5} q + 15p^{4} q^{2} + 20p^{3} q^{3} + 15p^{2} q^{4} + 6p q^{5} + q^{6}\)
(a) p(2 have valid licenses) = \(15p^{2} q^{4}\)
= \(15 (0.8)^{2} (0.2)^{4} = 15(0.64)(0.0016)\)
= \(0.01536 \approxeq 0.015\) (3 d.p)
(b) p(more than 3 have valid licenses) = \(p^{6} + 6p^{5} q + 15p^{4} q^{2}\)
= \((0.8)^{6} + 6(0.8)^{5} (0.2) + 15(0.8)^{4} (0.2)^{2}\)
= \(0.262144 + 0.393216 + 0.24578\)
= \(0.90112 \approxeq 0.901\) (3 d.p)
(c) p(at least 5 have valid licenses) = \(p^{6} + 6p^{5} q\)
= \((0.8)^{6} + 6(0.8)^{5} (0.2)\)
= \(0.262144 + 0.393216\)
= \(0.65536 \approxeq 0.655\).