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Three Persons A, B and C Apply for a Job of Manager in a Private Company. Chances of Their Selection (A, B and C) Are in Ratio 1 : 2 :4. , Find the Probability that It is Due to the Appointment of C.

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Question

Three persons A, B and C apply for a job of Manager in a Private Company. Chances of their selection (A, B and C) are in the ratio 1 : 2 :4. The probabilities that A, B and C can introduce changes to improve profits of the company are 0.8, 0.5 and 0.3, respectively. If the change does not take place, find the probability that it is due to the appointment of C.

Sum
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Solution

Let \[E_1 , E_2 \text{ and }  E_3\]  be the events denoting the selection of A, B and C as managers, respectively. \[P\left( E_1 \right)\] = Probability of selection of A = \[\frac{1}{7}\] \[P\left( E_2 \right)\]  = Probability of selection of B =\[\frac{2}{7}\] \[P\left( E_3 \right)\]  = Probability of selection of C = \[\frac{4}{7}\] Let A be the event denoting the change not taking place.

\[P\left( \frac{A}{E_1} \right)\]   = Probability that A does not introduce change = 0.2
\[P\left( \frac{A}{E_2} \right)\]  = Probability that B does not introduce change = 0.5
\[P\left( \frac{A}{E_3} \right)\] = Probability that C does not introduce change = 0.7
∴ Required probability = \[P\left( \frac{E_3}{A} \right)\]
By Bayes' theorem, we have
\[P\left( \frac{E_3}{A} \right)\]
\[ = \frac{P\left( E_3 \right)P\left( \frac{A}{E_3} \right)}{P\left( E_1 \right)P\left( \frac{A}{E_1} \right) + P\left( E_2 \right)P\left( \frac{A}{E_2} \right) + P\left( E_3 \right)P\left( \frac{A}{E_3} \right)}\]
\[ = \frac{\frac{4}{7} \times 0 . 7}{\frac{1}{7} \times 0 . 2 + \frac{2}{7} \times 0 . 5 + \frac{4}{7} \times 0 . 7}\]
\[ = \frac{2 . 8}{0 . 2 + 1 + 2 . 8}\]
\[ = \frac{2 . 8}{4} = 0 . 7\]
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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 24 | Page 97

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