Question

A student appears for test I, II and III. The student is successful if he passes either in tests I and II or test I and III. The probabilities of the student passing in tests I, II, III are p, q, and $\dfrac{1}{2}$ , respectively. If the probability that the student is successful is $\dfrac{1}{2}$ , then
(a) $p=1,q=0$
(b) $p=\dfrac{2}{3},q=\dfrac{1}{2}$
(c) $p=\dfrac{3}{5},q=\dfrac{2}{3}$
(d) there are infinitely many values of p and q

Answer 1

Hint: First, we will denote the probability of students passing in test I, II, III as $P\left( I \right),P\left( II \right),P\left( III \right)$. Then, we will use the concept of A union B formula i.e. $P\left( A\cup B \right)=P\left( A \right)+P\left( B \right)-P\left( A\cap B \right)$. So, we will be getting our equation as $P\left( S \right)=P\left( I\text{ and }II \right)+P\left( I\text{ and }III \right)-P\left( I\text{ and }II\text{ and }III \right)$ where S is probability of students successfully passed. So, on solving this equation we will get one equation at the end. So, we will then use the option method to check whether LHS is equal to RHS or not. Thus, we will get the answer.

Complete step-by-step answer:
Now, it is given that the probability of a student being successful is denoted as $P\left( S \right)=\dfrac{1}{2}$ . Also, a student is successful if he passes either in tests I and II or test I and III. So, in mathematical form it is written as
$P\left( S \right)=P\left( I\text{ and }II \right)+P\left( I\text{ and }III \right)-P\left( I\text{ and }II\text{ and }III \right)$
Here, ‘and’ means multiplication and ‘or’ means addition. This equation is similar probability of A union B given as $P\left( A\cup B \right)=P\left( A \right)+P\left( B \right)-P\left( A\cap B \right)$ .
So, we can write the equation as
$P\left( S \right)=P\left( I \right)P\left( II \right)+P\left( I \right)P\left( III \right)-P\left( I \right)P\left( II \right)P\left( III \right)$
Now, on substituting the values, we get
$\dfrac{1}{2}=pq+\left( p\cdot \dfrac{1}{2} \right)-\left( pq\cdot \dfrac{1}{2} \right)$
On further simplifying, we get
$\dfrac{1}{2}=pq+\dfrac{p}{2}-\dfrac{pq}{2}$
We will cancel the denominator term making it equal. So, we get
$1=2pq+p-pq$
$1=pq+p$
Taking p common from RHS, we get
$1=p\left( 1+q \right)$ ……………………………………(1)
Now, we will use the option method here, keeping the values of p and q and checking whether it is the same as the LHS side or not.
Taking option (a): $p=1,q=0$ . On substituting this value, we will get an equation as $1=1\left( 1+0 \right)=1$ . So, LHS is the same as RHS. So, this option is correct.
Taking option (b): $p=\dfrac{2}{3},q=\dfrac{1}{2}$ . On substituting this value, we will get equation as $1=\dfrac{2}{3}\left( 1+\dfrac{1}{2} \right)=\dfrac{2}{3}\cdot \dfrac{3}{2}=1$ . So, LHS is the same as RHS. So, this option is correct.
Taking option (c): $p=\dfrac{3}{5},q=\dfrac{2}{3}$ . On substituting this value, we will get equation as $1=\dfrac{3}{5}\left( 1+\dfrac{2}{3} \right)=\dfrac{3}{5}\cdot \dfrac{5}{3}=1$ . So, LHS is the same as RHS. So, this option is correct.
So, along with these three options there can be many infinite values of p and q. So, here all the options are correct.
Hence, option (a), (b), (c), (d) are correct.

Note: Another approach in solving this problem is by taking complement of the third event which is not in the passing criteria. We know that passing criteria is given as $P\left( S \right)=P\left( I \right)P\left( II \right)+P\left( I \right)P\left( III \right)$ so, here we will add the independent events of three test and also will find complement of event which s not given. So, we can write it as
$P\left( S \right)=P\left( I \right)P\left( II \right)\left( 1-P\left( III \right) \right)+P\left( I \right)P\left( III \right)\left( 1-P\left( II \right) \right)+P\left( I \right)P\left( II \right)P\left( III \right)$ . In solving this we will get our answer the same as done in the solution part.