Question

A and B are two candidates seeking admission to a college. The probability that A is selected is $P\left( A \right)=0.7$ and the probability that one of them is selected is 0.6. Find the probability of B getting selected?

Answer 1

Hint: Here for solving this question we will assume the probability of B getting selected as $p$. From the basic concept as we know that the total probability, in any case, is 1 that is the sum of the probability of an event happening and the complement of the probability of that event will be 1. We will use these 2 points and derive the answer using the given information.

Complete step-by-step solution
Given in the question that A and B are two candidates seeking admission in a college. The probability that A is selected is $P\left( A \right)=0.7$
The probability that one of them is selected is 0.6.
This can be mathematically written as $P\left( A \right)P\left( \overline{B} \right)+P\left( \overline{A} \right)P\left( B \right)=0.6$.
Let us assume that the probability of B getting selected is $p$. This can be mathematically written as $P\left( B \right)=p$.
As we know that the total probability, in any case, is 1 that is the sum of the probability of an event happening and the complement of the probability of that event will be 1.
This can be mathematically represented as the following for an event B: $P\left( B \right)+P\left( \overline{B} \right)=1$.
Since here we have $P\left( B \right)=p$ we can say that $P\left( \overline{B} \right)=1-p$.
By using these values $P\left( B \right)=p$ and $P\left( \overline{B} \right)=1-p$ and $P\left( A \right)=0.7$ in $P\left( A \right)P\left( \overline{B} \right)+P\left( \overline{A} \right)P\left( B \right)=0.6$.
We will have
$0.7\left( 1-p \right)+P\left( \overline{A} \right)p=0.6$
Since we know that $P\left( \overline{A} \right)+P\left( A \right)=1$ we can say that $P\left( \overline{A} \right)=1-0.7$ since it is given that $P\left( A \right)=0.7$ so we have $P\left( \overline{A} \right)=0.3$.
By using this value $P\left( \overline{A} \right)=0.3$ we can simplify this as
$\begin{align}
  & 0.7\left( 1-p \right)+0.3p=0.6 \\
 & \Rightarrow 0.7-0.7p+0.3p=0.6 \\
 & \Rightarrow 0.7-0.4p=0.6 \\
 & \Rightarrow 0.7-0.4p-0.6=0 \\
 & \Rightarrow 0.1-0.4p=0 \\
 & \Rightarrow 0.1=0.4p \\
 & \Rightarrow p=\dfrac{0.1}{0.4} \\
 & \Rightarrow p=\dfrac{1}{4} \\
 & \Rightarrow p=0.25 \\
\end{align}$
Here we now have the value of $p$ that is 0.25.
So we can say that $P\left( B \right)=0.25$ that is the probability of B getting selected for the admission in the college is 0.25.

Note: While solving questions of these type we should note that the probability that one of them is selected is mathematically written as $P\left( A \right)P\left( \overline{B} \right)+P\left( \overline{A} \right)P\left( B \right)$. Not as $P(A-B)+P(B-A)$ if by mistake we took this expression we will end up having a mess. Here 0.6 is the probability of selecting anyone, it can be A or B so do not assume this as the probability of selecting any particular person.