Probability 3 Question 18
18. A student appears for tests I, II and III. The student is successful if he passes either in tests I and II or tests I and III. The probabilities of the student passing in tests I, II and III are $p, q$ and $\frac{1}{2}$, respectively. If the probability that the student is successful, is $\frac{1}{2}$, then
(a) $p=q=1$
(b) $p=q=\frac{1}{2}$
(c) $p=1, q=0$
(d) $p=1, q=\frac{1}{2}$
(1986, 2M)
Show Answer
Solution:
- Let $A, B$ and $C$ denote the events of passing the tests I, II and III, respectively.
Evidently $A, B$ and $C$ are independent events.
According to given condition,
$$ \begin{aligned} \frac{1}{2} & =P[(A \cap B) \cup(A \cap C)] \\ & =P(A \cap B)+P(A \cap C)-P(A \cap B \cap C) \\ & =P(A) P(B)+P(A) \cdot P(C)-P(A) \cdot P(B) \cdot P(C) \\ & =p q+p \cdot \frac{1}{2}-p q \cdot \frac{1}{2} \\ \Rightarrow 1 & =2 p q+p-p q \quad \Rightarrow \quad 1=p(q+1) \end{aligned} $$
The values of option (c) satisfy Eq. (i).
[Infact, Eq. (i) is satisfied for infinite number of values of $p$ and $q$. If we take any values of $q$ such that $0 \leq q \leq 1$, then, $p$ takes the value $\frac{1}{q+1}$. It is evident that, $0<\frac{1}{q+1} \leq 1$ i.e. $0<p \leq 1$. But we have to choose correct answer from given ones.]
