A survey conducted revealed that four out of every twenty taxi drivers do not have a valid driving license. If 6 drivers are selected at random, calculate, correct to three decimal places, the probability that
(a) exactly 2 ;
(b) more than 3 ;
(c) at least 5; have valid driving license.
Explanation
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).