Q243 / 270 · NDA 2016 (Phase 1) · Mathematics

Two independent events A and B have $\mathrm{P}(\mathrm{A})=\frac{1}{3}$ and $\mathrm{P}(\mathrm{B})=\frac{3}{4}$. What is the probability that exactly one of the two events A or B occurs?

  • A
    $\frac{1}{4}$
  • B
    $\frac{5}{6}$
  • C
    $\frac{5}{12}$
  • Correct answer
    D
    $\frac{7}{12}$

Solution

$A$ and $B$ are independent. $P(A)=\frac{1}{3} \quad P(B)=\frac{3}{4}$ We want to find probability that exactly one of the two events $A$ or $B$ occurs i.e. when $A$ occurs $B$ does not and vice-versa. Lets take desired prob. is $P$. $\begin{aligned} \therefore P & =P(A)(1-P(B))+P(B)(1-P(A)) \\ & =\frac{1}{3}\left(1-\frac{3}{4}\right)+\frac{3}{4}\left(1-\frac{1}{3}\right) \\ & =\frac{1}{3} \times \frac{1}{4}+\frac{3 \times 2}{12} \\ P & =\frac{7}{12} \end{aligned}$

Hence, the correct answer is option: 'D'.

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