Probability 2 Question 4
4. For the three events $A, B$ and $C, P($ exactly one of the events $A$ or $B$ occurs $)=P$ (exactly one of the events $B$ or $C$ occurs $)=P($ exactly one of the events $C$ or $A$ occurs $)$ $=p$ and $P$ (all the three events occurs simultaneously) $=p^{2}$, where $0<p<\frac{1}{2}$. Then, the probability of atleast one of the three events $A, B$ and $C$ occurring is
(1996, 2M)
(a) $\frac{3 p+2 p^{2}}{2}$
(b) $\frac{p+3 p^{2}}{4}$
(c) $\frac{p+3 p^{2}}{2}$
(d) $\frac{3 p+2 p^{2}}{4}$
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Answer:
Correct Answer: 4. (c)
Solution:
- We know that,
$P$ (exactly one of $A$ or $B$ occurs)
$$ \begin{array}{ll} & =P(A)+P(B)-2 P(A \cap B) \\ \therefore & P(A)+P(B)-2 P(A \cap B)=p \\ \text { Similarly, } & P(B)+P(C)-2 P(B \cap C)=p \\ \text { and } & P(C)+P(A)-2 P(C \cap A)=p \end{array} $$
On adding Eqs. (i), (ii) and (iii), we get
$$ \begin{gathered} 2[P(A)+P(B)+P(C)-P(A \cap B) \\ -P(B \cap C)-P(C \cap A)]=3 p \\ \Rightarrow P(A)+P(B)+P(C)-P(A \cap B) \\ -P(B \cap C)-P(C \cap A)=\frac{3 p}{2} \quad \ldots(v) \end{gathered} $$
It also given that, $P(A \cap B \cap C)=p^{2}$
$\therefore \quad P$ (at least one of the events $A, B$, and $C$ occurs)
$$ \begin{aligned} & =P(A)+P(B)+P(C)-P(A \cap B) \\ & -P(B \cap C)-P(C \cap A)+P(A \cap B \cap C) \\ & =\frac{3 p}{2}+p^{2} \quad \text { [from Eqs. (iv) and (v)] } \\ & =\frac{3 p+2 p^{2}}{2} \end{aligned} $$
