Question

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

Answer 1

Hint: First of all, we are going to find the probability of selecting a manager from the three managers A, B and C. The ratio of their chances of selection is 4:1:2. Let us assume that the probability of selecting a manager A, B and C be the events ${{E}_{A}},{{E}_{B}},{{E}_{C}}$. Multiplying the ratio by x we will get the individual number for the three managers so the chance of selecting manager A, B and C is 4x, x and 2x respectively. to find the probability of selecting manager A, divide chance of selecting A to chance of selecting all the three managers. Similarly, find the probability for managers B and C as well. Now, let us denote event E is the event when a radical change occurs so probability that a radical change occurs due to manager A is equal to $P\left( \dfrac{E}{{{E}_{A}}} \right)$ which is given as 0.3 in the above question. Similarly, we can write for manager B and C. To find the probability that if a change occurs, it will be due to manager B is calculated using Baye’s theorem $P\left( \dfrac{{{E}_{B}}}{E} \right)=\dfrac{P\left( {{E}_{B}} \right)P\left( \dfrac{E}{{{E}_{B}}} \right)}{P\left( {{E}_{A}} \right)P\left( \dfrac{E}{{{E}_{A}}} \right)+P\left( {{E}_{B}} \right)P\left( \dfrac{E}{{{E}_{B}}} \right)+P\left( {{E}_{C}} \right)P\left( \dfrac{E}{{{E}_{C}}} \right)}$ . Substitute the values that we have described above and get the answer.

Complete step by step answer:
We have given the ratio of chances of selecting manager A, B and C as 4:1:2. We are going to convert the ratio into a number by multiplying with x so the chances of selecting A, B and C are as follows:
Let us assume that the events of selecting manager A, B and C are ${{E}_{A}},{{E}_{B}}\And {{E}_{C}}$ respectively.
Chances of selecting A $=4x$
Chances of selecting B $=x$
Chances of selecting C $=2x$
Total chance of selecting all the mangers is the addition of chance of selecting A, B and C individually.
$\begin{align}
  & 4x+x+2x \\
 & =7x \\
\end{align}$
Probability of selecting manager A is equal to:
$\begin{align}
  & P\left( {{E}_{A}} \right)=\dfrac{{{E}_{A}}}{{{E}_{A}}+{{E}_{B}}+{{E}_{C}}} \\
 & \Rightarrow P\left( {{E}_{A}} \right)=\dfrac{4x}{7x}=\dfrac{4}{7} \\
\end{align}$
Probability of selecting manager B is equal to:
$\begin{align}
  & P\left( {{E}_{B}} \right)=\dfrac{{{E}_{B}}}{{{E}_{A}}+{{E}_{B}}+{{E}_{C}}} \\
 & \Rightarrow P\left( {{E}_{B}} \right)=\dfrac{x}{7x}=\dfrac{1}{7} \\
\end{align}$
Probability of selecting manager C is equal to:
$\begin{align}
  & P\left( {{E}_{C}} \right)=\dfrac{{{E}_{C}}}{{{E}_{A}}+{{E}_{B}}+{{E}_{C}}} \\
 & \Rightarrow P\left( {{E}_{C}} \right)=\dfrac{2x}{7x}=\dfrac{2}{7} \\
\end{align}$
It is given that the probabilities that the radical change occurs due to manager A, B and C are:
$\begin{align}
  & P\left( \dfrac{E}{{{E}_{A}}} \right)=0.3 \\
 & P\left( \dfrac{E}{{{E}_{B}}} \right)=0.8 \\
 & P\left( \dfrac{E}{{{E}_{C}}} \right)=0.5 \\
\end{align}$
Now, to find the probability that if a change has been occurred, it is due to the appointment of B which we are going to calculate by Baye’s theorem,
$P\left( \dfrac{{{E}_{B}}}{E} \right)=\dfrac{P\left( {{E}_{B}} \right)P\left( \dfrac{E}{{{E}_{B}}} \right)}{P\left( {{E}_{A}} \right)P\left( \dfrac{E}{{{E}_{A}}} \right)+P\left( {{E}_{B}} \right)P\left( \dfrac{E}{{{E}_{B}}} \right)+P\left( {{E}_{C}} \right)P\left( \dfrac{E}{{{E}_{C}}} \right)}$
Substituting the probabilities in R.H.S of the above equation which we have calculated above we get,
$P\left( \dfrac{{{E}_{B}}}{E} \right)=\dfrac{\dfrac{1}{7}(0.8)}{\dfrac{4}{7}(0.3)+\dfrac{1}{7}(0.8)+\dfrac{2}{7}(0.5)}$
In the R.H.S of the above equation, $\dfrac{1}{7}$ will be cancelled out from the numerator and the denominator then the equation which we are left with:
$\begin{align}
  & P\left( \dfrac{{{E}_{B}}}{E} \right)=\dfrac{0.8}{4\left( 0.3 \right)+0.8+2\left( 0.5 \right)} \\
 & \Rightarrow P\left( \dfrac{{{E}_{B}}}{E} \right)=\dfrac{0.8}{1.2+0.8+1.0} \\
 & \Rightarrow P\left( \dfrac{{{E}_{B}}}{E} \right)=\dfrac{0.8}{3.0} \\
 & \Rightarrow P\left( \dfrac{{{E}_{B}}}{E} \right)=\dfrac{8}{30}=\dfrac{4}{15} \\
\end{align}$

Hence, the probability that if a change occurs it will be due to the appointment of manager B is equal to $\dfrac{4}{15}$.

Note: The question demands the knowledge of the Baye’s theorem, if you don’t know the Baye’s theorem then you cannot solve this problem. And one more thing the possibility of making calculation mistakes is pretty high here. There is a general way to check any probability is that the probability of any event is always lying between 0 and 1 including 0 and 1. If your probability is greater than 1 or less than 0 that definitely means that you are getting the wrong probability.