Q264 / 270 · NDA 2016 (Phase 2) · Mathematics

Three independent events, $\mathrm{A}_1, \mathrm{~A}_2$ and $\mathrm{A}_3$ occur with probabilities $\mathrm{P}\left(\mathrm{A}_{\mathrm{i}}\right)=\frac{1}{1+\mathrm{i}}, \mathrm{i}=1,2,3$. What is the probability that at least one of the three events occurs?

  • A
    $\frac{1}{4}$
  • B
    $\frac{2}{3}$
  • Correct answer
    C
    $\frac{3}{4}$
  • D
    $\frac{1}{24}$

Solution

$\begin{aligned} & P\left(A_1\right)=\frac{1}{1+1}=\frac{1}{2} \\ & P\left(A_2\right)=\frac{1}{3} \\ & P\left(A_3\right)=\frac{1}{4}\end{aligned}$ $\therefore \quad$ Probability that at least one of these events occur is $P\left(A_1 \cup A_2 \cup A_3\right)$. Also $A_1, A_2 \& A_3$ are independent events. $\begin{aligned} P\left(A_1 \cup A_2 \cup A_3\right)= & P\left(A_1\right)+P\left(A_2\right)+P\left(A_3\right) \\ & -P\left(A_1 \cap A_2\right)-P\left(A_1 \cap A_3\right) \\ & -P\left(A_2 \cap A_3\right)+P\left(A_1 \cap A_2 \cap A_3\right) \\ =\frac{1}{2}+ & +\frac{1}{3}+\frac{1}{4}-\left(\frac{1}{2} \times \frac{1}{3}\right)-\left(\frac{1}{2} \times \frac{1}{4}\right) \\ & -\left(\frac{1}{3} \times \frac{1}{4}\right)+\left(\frac{1}{2} \times \frac{1}{3} \times \frac{1}{4}\right) \\ = & \frac{3}{4} . \end{aligned}$

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

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