Question

For a post, three persons A, B, and C appear in the interview. The probability of A being selected is twice that of B and the probability of B being selected is thrice that of C, what are the individual probabilities of A, B, C being selected?
(A) \[P\left( {{E}_{1}} \right)=\dfrac{3}{5},P\left( {{E}_{2}} \right)=\dfrac{1}{10},P\left( {{E}_{3}} \right)=\dfrac{3}{10}\]
(B) \[P\left( {{E}_{1}} \right)=\dfrac{3}{10},P\left( {{E}_{2}} \right)=\dfrac{3}{5},P\left( {{E}_{3}} \right)=\dfrac{1}{10}\]
(C) \[P\left( {{E}_{1}} \right)=\dfrac{3}{10},P\left( {{E}_{2}} \right)=\dfrac{3}{10},P\left( {{E}_{3}} \right)=\dfrac{1}{10}\]
(D) \[P\left( {{E}_{1}} \right)=\dfrac{3}{5},P\left( {{E}_{2}} \right)=\dfrac{3}{10},P\left( {{E}_{3}} \right)=\dfrac{1}{10}\]

Answer 1

Hint: Assume that the probability of the selection of person c for the post is x. It is given that the probability of selection of person B is thrice of the probability of selection of person C and the probability of selection of person A is twice the probability of selection of person B. So, the probabilities of selection of person A and person B are \[6x\] and \[3x\] respectively. We know that the summation of the probabilities of all outcomes is equal to 1. So, The probability of selection of person A + The probability of selection of person B + The probability of selection of person C = 1. Now, solve it further to get the value of x. Using the value of x, get the value of the probabilities of selection of person A, person B, and person C.

Complete step-by-step answer:
First of all, let us assume that the probability of the selection of person c for the post is x.
The probability of selection of person C, \[P\left( {{E}_{3}} \right)\] = \[x\] ……………………...(1)
It is given that the probability of selection of person B is thrice of the probability of selection of person C.
From equation (1), we have the probability of selection of person C.
The probability of selection of person B, \[P\left( {{E}_{2}} \right)\] = \[x\times 3=3x\] ……………………………(2)
It is also given that the probability of selection of person A is twice the probability of selection of person B.
From equation (2), we have the probability of selection of person B.
The probability of selection of person A, \[P\left( {{E}_{1}} \right)\] = \[2\times 3x=6x\] ……………………………..(3)
We know that the summation of the probabilities of all outcomes is equal to 1. So,
The probability of selection of person A + The probability of selection of person B + The probability of selection of person C = 1 …………………………………..(4)
From equation (1), equation (2), and equation (3) we have the probability of selection of person C, person B, and person A respectively.
Now, from equation (1), equation (2), equation (3) and equation (4), we get
\[\begin{align}
  & \Rightarrow P\left( {{E}_{3}} \right)+P\left( {{E}_{2}} \right)+P\left( {{E}_{1}} \right)=1 \\
 & \Rightarrow x+3x+6x=1 \\
 & \Rightarrow 10x=1 \\
\end{align}\]
\[\Rightarrow x=\dfrac{1}{10}\] ……………………….(5)
Now, putting the value of x in equation (1), we get
The probability of selection of person C, \[P\left( {{E}_{3}} \right)\] = \[x=\dfrac{1}{10}\] .
Now, putting the value of x in equation (2), we get
The probability of selection of person B, \[P\left( {{E}_{2}} \right)\] = \[3x=3\times \dfrac{1}{10}=\dfrac{3}{10}\] .
Now, putting the value of x in equation (3), we get
The probability of selection of person A, \[P\left( {{E}_{1}} \right)\] = \[6x=6\times \dfrac{1}{10}=\dfrac{3}{5}\] .
Therefore, the probabilities of selection of person A, person B, and person C are \[P\left( {{E}_{1}} \right)=\dfrac{3}{5}\] , \[P\left( {{E}_{2}} \right)=\dfrac{3}{10}\] , and \[P\left( {{E}_{3}} \right)=\dfrac{1}{10}\] .
Hence, the correct option is (D).

Note: We can also solve this question, by observing all the options given. It is given that the probability of A being selected is twice that of B and the probability of B being selected is thrice that of C. In option (D), we can see that \[\dfrac{3}{10}\] is thrice of \[\dfrac{1}{10}\] i.e, the probability of selection of B is thrice of the probability of selection of C and \[\dfrac{3}{5}\] is twice of \[\dfrac{3}{10}\] i.e, the probability of selection of A is twice of the probability of selection of B. Therefore, option (D) satisfies the given information. Hence, option (D) is the correct one.