(a) An association is made up of 6 farmers and 8 traders. If an executive body of 4 members is to be formed, find the probability that it will consist of at least two farmers. (b) The probability of an accident occurring in a given month in factories X, Y, and Z are \(\frac{1}{5}, \frac{1}{12} \) and \(\frac{1}{6}\) respectively.
Find the probability that the accident will occur in:
i) none of the factories;
(ii) all the factories;
(iii) at least one factory.
Find the probability that the accident will occur in:
i) none of the factories;
(ii) all the factories;
(iii) at least one factory.
Take Free Practice Test On 2026 JAMB UTME, Post UTME, WAEC SSCE, GCE, NECO SSCE
Correct Answer: Option
Explanation:
No. of farmers = 6
No. of traders = 8
Total = 14
Pr(at least two farmers)
= \(\frac{^6C_2 \times ^8C_2}{^{14}C_4} + \frac{^6C_3 \times ^8C_1}{^{14}}+ \frac{^6C_4 \times ^8C_0}{^{14}C_4}\)
= \(\frac{15 \times 28}{1001} + \frac{20 \times 8}{1001} + \frac{15 \times 1}{1001}\)
= \(\frac{420}{1001} + \frac{160}{1001} + \frac{15}{1001}\)
= \(\frac{420 + 160 + 15}{1001}\)
= \(\frac{595}{1001}\)
(b) P(x) = \(\frac{1}{5}\)
p(x\(^1\)) = 1 - \(\frac{1}{5}\) = \(\frac{4}{5}\)
p(y) = \(\frac{1}{2}\), p(y\(^1\))
= 1 - \(\frac{1}{12}\)
= \(\frac{11}{12}\)
p(2) = \(\frac{1}{6}\)
p(z\(^1\)) = 1 - \(\frac{1}{6}\)
= \(\frac{5}{6}\)
(i) Pr(None)
= \(\frac{4}{5} \times \frac{11}{12} \times \frac{5}{6} = \frac{11}{18}\)
(ii) Pr(all the factories)
= \(\frac{1}{5} \times \frac{1}{12} \times \frac{1}{6} = \frac{1}{360^o}\)
(iii) Pr (at least one factory)
= 1 - Pr (none) = \(\frac{1}{1} - \frac{11}{18}\)
= \(\frac{18 - 11}{18}\)
= \(\frac{7}{18}\)
No. of farmers = 6
No. of traders = 8
Total = 14
Pr(at least two farmers)
= \(\frac{^6C_2 \times ^8C_2}{^{14}C_4} + \frac{^6C_3 \times ^8C_1}{^{14}}+ \frac{^6C_4 \times ^8C_0}{^{14}C_4}\)
= \(\frac{15 \times 28}{1001} + \frac{20 \times 8}{1001} + \frac{15 \times 1}{1001}\)
= \(\frac{420}{1001} + \frac{160}{1001} + \frac{15}{1001}\)
= \(\frac{420 + 160 + 15}{1001}\)
= \(\frac{595}{1001}\)
(b) P(x) = \(\frac{1}{5}\)
p(x\(^1\)) = 1 - \(\frac{1}{5}\) = \(\frac{4}{5}\)
p(y) = \(\frac{1}{2}\), p(y\(^1\))
= 1 - \(\frac{1}{12}\)
= \(\frac{11}{12}\)
p(2) = \(\frac{1}{6}\)
p(z\(^1\)) = 1 - \(\frac{1}{6}\)
= \(\frac{5}{6}\)
(i) Pr(None)
= \(\frac{4}{5} \times \frac{11}{12} \times \frac{5}{6} = \frac{11}{18}\)
(ii) Pr(all the factories)
= \(\frac{1}{5} \times \frac{1}{12} \times \frac{1}{6} = \frac{1}{360^o}\)
(iii) Pr (at least one factory)
= 1 - Pr (none) = \(\frac{1}{1} - \frac{11}{18}\)
= \(\frac{18 - 11}{18}\)
= \(\frac{7}{18}\)