# For three persons A, B, C the chances of being selected as manager of a firm are in the ratio 4:1:2 respectively. The respective probabilities for them to introduce radical changes in market strategy are 0.3, 0.8 and 0.5. If the changes take place, find the probability that

a) It was due to the appointment of A.

b) It was due to appointment of B or C.

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**Hint:**This problem of probability is based on the Bayes’ Theorem. We are to use Bayes’ Theorem to solve this question.

**Complete step-by-step answer:**

We are given that for three persons A, B, C the chances of being selected as manager of a firm are in the ratio 4:1:2 respectively. The respective probabilities for them to introduce radical changes in market strategy are 0.3, 0.8 and 0.5.

At first we will start by explaining Bayes’ Theorem.

In probability theory and statistics, Bayes’ theorem (alternatively Bayes’ law or Bayes’ rule) describes the probability of an event, based on prior knowledge of conditions that might be related to the event.

$P({A_i}|B) = \dfrac{{P(B|{A_i})*P({A_i})}}{{\sum\limits_{i = 1}^n {P(B|{A_i})*P({A_i})} }}$

As in this question the probability of introducing radical changes in the market is based on the probability of them (A, B, and C) to get selected first. That is without getting selected as manager one cannot make changes in the strategy.

(a) Let us find probabilities for selecting the members first.

P (A) for getting selected = $\dfrac{4}{{4 + 1 + 2}}$=$\dfrac{4}{7}$

P (B) for getting selected = $\dfrac{1}{{4 + 1 + 2}}$=$\dfrac{1}{7}$

P (C) for getting selected = $\dfrac{2}{{4 + 1 + 2}}$=$\dfrac{2}{7}$

$P({A_i}|D) = \dfrac{{P(D|{A_i})*P({A_i})}}{{\sum\limits_{i = 1}^n {P(D|{A_i})*P({A_i})} }}$

Where,$D$ is the event of a member to make changes in the strategy.

And ${A_i}$is the event of selection of a member.

$P({A_1}|D) = \dfrac{{P(D|{A_1})*P({A_1})}}{{\sum\limits_{i = 1}^3 {P(D|{A_i})*P({A_i})} }}$

Here, ${A_1}$ is $A$, ${A_2}$ is $B$, and ${A_3}$ is $C$.

$P({A_1}|D) = \dfrac{{0.3 \times \dfrac{4}{7}}}{{0.3 \times \dfrac{4}{7} + 0.8 \times \dfrac{1}{7} + 0.5 \times \dfrac{2}{7}}}$

On simplifying this we get,

$P({A_1}|D) = 0.4$

(b) Now for the second part it is asked to calculate probability for either B or C. This can calculated as the probability of getting either B or C will be equal to total probability of getting anything except A. Thus,

\[P\left( {B{\text{ }}or{\text{ }}C} \right)\] = $1 - 0.4 \Rightarrow 0.6$

Thus, the answer is $P({A_1}|D) = 0.4$ and \[P\left( {B{\text{ }}or{\text{ }}C} \right)\]$ = 0.6$.

**Note:**Bayes’ theorem is bound to condition that the second event is based on prior knowledge of conditions that might be related to the first event. That is, it must not be applied when the two events are independent to each other.