Question

A and B are two candidates seeking admission in I.I.T. The probability that A is selected is \[0.5\] and the probability that both A and B are selected is at most \[0.3\] . Is it possible that the probability of B getting selected is \[0.9\] ?

Answer 1

Hint: To check the possibility that the probability of B getting selected is \[0.9\] , we will use the identity $p(A\cup B)=p(A)+p(B)-p(A\cap B)$ . Substitute the given values in this equation and also use the concept that $p(A\cup B)\le 1$ . By proper substitution and simplification, we can get the probability of B getting selected.

Complete step by step answer:
We need to check the possibility that the probability of B getting selected is \[0.9\] .
Given that the probability that A is selected is \[0.5\] . That is
$p(A)=0.5$ .
Let $p(B)$ denote the probability that B is selected.
Also it is given that the probability that both A and B are selected is at most \[0.3\] .
That is, \[p(A\cap B)\le 0.3...(i)\] .
We know that $p(A\cup B)=p(A)+p(B)-p(A\cap B)$
Rearranging this, we will get
$p(A\cap B)=p(A)+p(B)-p(A\cup B)$
Using $(i)$ , we can write the above equation has
$p(A)+p(B)-p(A\cup B)\le 0.3$
Substituting the value of $p(A)$ in the above equation, we get
$0.5+p(B)-p(A\cup B)\le 0.3$
From this $p(B)$ can be given as
$p(B)\le -0.2+p(A\cup B)...(a)$
We know that $p(A\cup B)\le 1$
Adding $-0.2$ to both the sides, we get
$-0.2+p(A\cup B)\le 1-0.2$
$\Rightarrow -0.2+p(A\cup B)\le 0.8$
Substituting the above equation in $(a)$ , we get
$p(B)\le 0.8$

Hence, it is not possible that the probability of B getting selected is \[0.9\].

Note: Probability rules need to be thoroughly studied to solve these problems. The main equations are $p(A\cap B)=p(A)+p(B)-p(A\cup B)$ and $p(A\cup B)\le 1$ . Be careful with the inequalities. There can be a possibility of error when the inequalities are not applied properly.