The probability of Jide, Atu and Obu solving a given problem are \(\frac{1}{12}\), \(\frac{1}{6}\) and \(\frac{1}{8}\) respectively. Calculate the probability that only one solves the problem.
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Correct Answer: Option D
Explanation:
\(P(Jide) = \frac{1}{12}; P(\text{not Jide}) = \frac{11}{12}\)
\(P(Atu) = \frac{1}{6}; P(\text{not Atu}) = \frac{5}{6}\)
\(P(Obu) = \frac{1}{8}; P(\text{not Obu}) = \frac{7}{8}\)
\(P(\text{only one of them}) = P(\text{Jide not Atu not Obu}) + P(\text{Atu not Jide not Obu}) + P(\text{Obu not Jide not Atu})\)
= \((\frac{1}{12} \times \frac{5}{6} \times \frac{7}{8}) + (\frac{1}{6} \times \frac{11}{12} \times \frac{7}{8}) + (\frac{1}{8} \times \frac{11}{12} \times \frac{5}{6})\)
= \(\frac{35}{576} + \frac{77}{576} + \frac{55}{576}\)
= \(\frac{167}{576}\)
\(P(Jide) = \frac{1}{12}; P(\text{not Jide}) = \frac{11}{12}\)
\(P(Atu) = \frac{1}{6}; P(\text{not Atu}) = \frac{5}{6}\)
\(P(Obu) = \frac{1}{8}; P(\text{not Obu}) = \frac{7}{8}\)
\(P(\text{only one of them}) = P(\text{Jide not Atu not Obu}) + P(\text{Atu not Jide not Obu}) + P(\text{Obu not Jide not Atu})\)
= \((\frac{1}{12} \times \frac{5}{6} \times \frac{7}{8}) + (\frac{1}{6} \times \frac{11}{12} \times \frac{7}{8}) + (\frac{1}{8} \times \frac{11}{12} \times \frac{5}{6})\)
= \(\frac{35}{576} + \frac{77}{576} + \frac{55}{576}\)
= \(\frac{167}{576}\)