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For A, B and C the Chances of Being Selected as the Manager of a Firm Are in the Ratio 4:1:2 Respectively. Find the Probability that It is Due to the Appointment of B Or C.

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Question

For AB and C the chances of being selected as the manager of a firm are in the ratio 4:1:2 respectively. The respective probabilities for them to introduce a radical change in marketing strategy are 0.3, 0.8 and 0.5. If the change does take place, find the probability that it is due to the appointment of B or C.

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Solution

Let AE1E2 and E3 denote the events that the change takes place, A is selected, B is selected and C is selected, respectively.

\[\therefore P\left( E_1 \right) = \frac{4}{7}\]
\[ P\left( E_2 \right) = \frac{1}{7} \]
\[ P\left( E_3 \right) = \frac{2}{7}\]
\[\text{ Now} , \]
\[P\left( A/ E_1 \right) = 0 . 3\]
\[P\left( A/ E_2 \right) = 0 . 8\]
\[P\left( A/ E_3 \right) = 0 . 5\]
\[\text{ Using Bayes' theorem, we get } \]
\[ P\left( E_1 /A \right) = \frac{P\left( E_1 \right)P\left( A/ E_1 \right)}{P\left( E_1 \right)P\left( A/ E_1 \right) + P\left( E_2 \right)P\left( A/ E_2 \right) + P\left( E_3 \right)P\left( A/ E_3 \right)}\]
\[ = \frac{\frac{4}{7} \times 0 . 3}{\frac{4}{7} \times 0 . 3 + \frac{1}{7} \times 0 . 8 + \frac{2}{7} \times 0 . 5}\]
\[ = \frac{1 . 2}{1 . 2 + 0 . 8 + 1} = \frac{1 . 2}{3} = \frac{12}{30} = \frac{2}{5}\]
\[ \therefore \text{ Required probability} = 1 - P\left( E_1 /A \right) = 1 - \frac{2}{5} = \frac{3}{5}\]

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Chapter 30: Probability - Exercise 31.7 [Page 97]

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R.D. Sharma Mathematics Volume 1 and 2 [English] Class 12
Chapter 30 Probability
Exercise 31.7 | Q 23 | Page 97

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