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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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प्रश्न

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.

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उत्तर

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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अध्याय 30: Probability - Exercise 31.7 [पृष्ठ ९७]

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आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
अध्याय 30 Probability
Exercise 31.7 | Q 24 | पृष्ठ ९७

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