Question

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 simultaneosuly) $=\frac{1}{16},$ then the probability that atleast one of the events occurs is ?

• A. $\frac{7}{32}$
• B. $\frac{7}{16}$
• C. $\frac{7}{64}$
• D. $\frac{3}{16}$

Answer 1

The correct option is B $\frac{7}{16}$

We have, P(exactly one of A or B occurs)
$=P(A\cup B)-P(A\cap B)$
$=P\left(A\right)+P\left(B\right)-2P(A\cap B)$
According to the question,
$P\left(A\right)+P\left(B\right)-2P(A\cap B)=\frac{1}{4}$ . . . (i)
$P\left(B\right)+P\left(C\right)-2P(B\cap C)=\frac{1}{4}$ . . . (ii)
and $P\left(C\right)+P\left(A\right)-2P(C\cap A)=\frac{1}{4}$ . . . (iii)
On adding Eqs. (i),(ii) and (iii), we get
$2\left[P\right(A)+P(B)+P(C)-P(A\cap B)-P(B\cap C)-P(C\cap A\left)\right]=\frac{3}{4}$
$\Rightarrow P\left(A\right)+P\left(B\right)+P\left(C\right)-P(A\cap B)-P(B\cap C)-P(C\cap A)=\frac{3}{8}$
$\therefore$ P(atleast one event occurs)
$=P(A\cup B\cup C)=P\left(A\right)+P\left(B\right)+P\left(C\right)-P(A\cap B)-P(B\cap C)-P(C\cap A)+P(A\cap B\cap C)=\frac{3}{8}+\frac{1}{16}=\frac{7}{16}(\because P(A\cap B\cap C)=\frac{1}{16})$