Question

The Probability of A, B, C solving a problem are $\dfrac{1}{3},\dfrac{2}{7},\dfrac{3}{8}$ respectively. If all three solve the problem simultaneously, the probability that exactly one of them will solve it.
A. $\dfrac{{25}}{{168}}$
B. $\dfrac{{25}}{{56}}$
C. $\dfrac{{20}}{{168}}$
D. $\dfrac{{30}}{{168}}$

Answer 1

In this problem, we have to find the probabilities such that when one of them is able to solve the remaining must fail in this way we will need to make 3 separate cases to solve each one of them and then add them all to get the final answer.

Complete step-by-step answer:
Case I:
Probability of A solving (remaining fail)
This will be given by $P(A) \times P\left( {{B^C}} \right) \times P\left( {{C^C}} \right)$
Now we know that

P(A) = \dfrac{1}{3}\\\ P\left( {{B^C}} \right) = \left( {1 - \dfrac{2}{7}} \right) = \dfrac{{7 - 2}}{7} = \dfrac{5}{7}\\\ P\left( {{C^C}} \right) = \left( {1 - \dfrac{3}{8}} \right) = \dfrac{{8 - 3}}{8} = \dfrac{5}{8} \end{array}$$ Which means that $$P(A) \times P\left( {{B^C}} \right) \times P\left( {{C^C}} \right) = \dfrac{1}{3} \times \dfrac{5}{7} \times \dfrac{5}{8} = \dfrac{{25}}{{168}}$$ Case II: Probability of B solving (remaining fail) This will be given by $$P\left( {{A^C}} \right) \times P\left( B \right) \times P\left( {{C^C}} \right)$$ Now we know that $$\begin{array}{l} P\left( {{A^C}} \right) = \left( {1 - \dfrac{1}{3}} \right) = \dfrac{{3 - 1}}{3} = \dfrac{2}{3}\\\ P\left( B \right) = \dfrac{2}{7}\\\ P\left( {{C^C}} \right) = \left( {1 - \dfrac{3}{8}} \right) = \dfrac{{8 - 3}}{8} = \dfrac{5}{8}\\\ \therefore P\left( {{A^C}} \right) \times P\left( B \right) \times P\left( {{C^C}} \right) = \dfrac{2}{3} \times \dfrac{2}{7} \times \dfrac{5}{8} = \dfrac{{20}}{{168}} \end{array}$$ Case III: Probability of C solving (remaining fail) This will be given by $$P\left( {{A^C}} \right) \times P\left( {{B^C}} \right) \times P\left( C \right)$$ Now we know that $$\begin{array}{l} P\left( {{A^C}} \right) = \left( {1 - \dfrac{1}{3}} \right) = \dfrac{{3 - 1}}{3} = \dfrac{2}{3}\\\ P\left( {{B^C}} \right) = \left( {1 - \dfrac{2}{7}} \right) = \dfrac{{7 - 2}}{7} = \dfrac{5}{7}\\\ P\left( C \right) = \dfrac{3}{8}\\\ \therefore P\left( {{A^C}} \right) \times P\left( {{B^C}} \right) \times P\left( C \right) = \dfrac{2}{3} \times \dfrac{5}{7} \times \dfrac{3}{8} = \dfrac{{30}}{{168}} \end{array}$$ **So, the correct answer is “Option D”.** **Note:** Here Students often get confused about when to add and when to multiply. They often multiply in between the cases and add all the probability individually which is a wrong concept.