Probability 2 Question 1
1. For three events $A, B$ and $C$, if $P$ (exactly one of $A$ or $B$ occurs $)=P$ (exactly one of $B$ or $C$ occurs $)=P$ (exactly one of $C$ or $A$ occurs $)=\frac{1}{4}$ and $P$ (all the three events occur simultaneously) $=\frac{1}{16}$, then the probability that atleast one of the events occurs, is
(a) $\frac{7}{32}$
(b) $\frac{7}{16}$
(c) $\frac{7}{64}$
(d) $\frac{3}{16}$
(2017 Main)
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Answer:
Correct Answer: 1. (b)
Solution:
- We have, $P$ (exactly one of $A$ or $B$ occurs)
$$ \begin{aligned} & =P(A \cup B)-P(A \cap B) \\ & =P(A)+P(B)-2 P(A \cap B) \end{aligned} $$
According to the question,
$$ \begin{aligned} & P(A)+P(B)-2 P(A \cap B)=\frac{1}{4} \\ & P(B)+P(C)-2 P(B \cap C)=\frac{1}{4} \\ & P(C)+P(A)-2 P(C \cap A)=\frac{1}{4} \end{aligned} $$
and
On adding Eqs. (i), (ii) and (iii), we get
$$ \begin{aligned} & 2[P(A)+P(B)+P(C)-P(A \cap B)-P(B \cap C) \\ & -P(C \cap A)]=\frac{3}{4} \\ & \Rightarrow P(A)+P(B)+P(C)-P(A \cap B)-P(B \cap C) \\ & -P(C \cap A)=\frac{3}{8} \\ & =P(A)+P(B)+P(C)-P(A \cap B)-P(B \cap C) \\ & -P(C \cap A)+P(A \cap B \cap C) \\ & =\frac{3}{8}+\frac{1}{16}=\frac{7}{16} \quad \because P(A \cap B \cap C)=\frac{1}{16} \end{aligned} $$
