(a) In a school, the ratio of those who passed to those who failed in a History test is 4 : 1. If 7 students are selected at random from the school, find, correct to two decimal places, the probability that :
(i) at least 3 passed the test ; (ii) between 3 and 6 students failed the test.
(b) A fair die is thrown five times; find the probability of obtaining a six three times.
Explanation
(a) Passed : Failed = 4 : 1
\(p(Passed) = \frac{4}{5} ; p(Failed) = \frac{1}{5}\)
(i) p(at least 3 passed) = 1 - [p(0) + p(1) + p(2)]
p(0 passed) = \(^{7}C_{0} (0.8)^{0} (0.2)^{7} = 0.0000128\)
p(1 passed) = \(^{7}C_{1} (0.8)^{1} (0.2)^{6} = 0.000358\)
p(2 passed) = \(^{7}C_{2} (0.8)^{2} (0.2)^{5} = 0.0043\)
p(at least 3 passed) = \(1 - [0.0000128 + 0.000358 + 0.0043] = 0.9953\)
(ii) p(between 3 and 6 students failed) = p(4 or 5 students failed)
= \((^{7}C_{4} (0.8)^{3} (0.2)^{4}) + (^{7}C_{5} (0.8)^{2} (0.2)^{5})\)
= \(0.02867 + 0.0043008\)
= \(0.03297\)
(b) p(a six) = p = \(\frac{1}{6}\), p(not a six) = q = \(\frac{5}{6}\)
Die thrown five times;
Binomial distribution function is \((p + q)^{5} = p^{5} + 5p^{4}q + 10p^{3} q^{2} + 10p^{2} q^{3} + 5pq^{4} + q^{5}\)
p(a six three times) = \(10p^{3} q^{2}\)
= \(10(\frac{1}{6})^{3} (\frac{5}{6})^{2}\)
= \(\frac{250}{7776}\)
= 0.03215