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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?
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that at least one of the three events occurs?
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Verified Answer
The correct answer is:
$\frac{3}{4}$
$P\left(A_{1}\right)=\frac{1}{1+1}=\frac{1}{2}$
$P\left(A_{2}\right)=\frac{1}{3}$
$P\left(A_{3}\right)=\frac{1}{4}$
$\therefore \quad$ 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.
$\begin{aligned} P\left(A_{1} \cup A_{2} \cup A_{3}\right)=& P\left(A_{1}\right)+P\left(A_{2}\right)+P\left(A_{3}\right) \\ &-P\left(A_{1} \cap A_{2}\right)-P\left(A_{1} \cap A_{3}\right) \\ &-P\left(A_{2} \cap A_{3}\right)+P\left(A_{1} \cap A_{2} \cap A_{3}\right) \\=\frac{1}{2} &+\frac{1}{3}+\frac{1}{4}-\left(\frac{1}{2} \times \frac{1}{3}\right)-\left(\frac{1}{2} \times \frac{1}{4}\right) \\ &-\left(\frac{1}{3} \times \frac{1}{4}\right)+\left(\frac{1}{2} \times \frac{1}{3} \times \frac{1}{4}\right) \\=& \frac{3}{4} . \end{aligned}$
$P\left(A_{2}\right)=\frac{1}{3}$
$P\left(A_{3}\right)=\frac{1}{4}$
$\therefore \quad$ 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.
$\begin{aligned} P\left(A_{1} \cup A_{2} \cup A_{3}\right)=& P\left(A_{1}\right)+P\left(A_{2}\right)+P\left(A_{3}\right) \\ &-P\left(A_{1} \cap A_{2}\right)-P\left(A_{1} \cap A_{3}\right) \\ &-P\left(A_{2} \cap A_{3}\right)+P\left(A_{1} \cap A_{2} \cap A_{3}\right) \\=\frac{1}{2} &+\frac{1}{3}+\frac{1}{4}-\left(\frac{1}{2} \times \frac{1}{3}\right)-\left(\frac{1}{2} \times \frac{1}{4}\right) \\ &-\left(\frac{1}{3} \times \frac{1}{4}\right)+\left(\frac{1}{2} \times \frac{1}{3} \times \frac{1}{4}\right) \\=& \frac{3}{4} . \end{aligned}$
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