6

All events are independent.

$$\Pr(A) = \frac{9}{10}$$ $$\Pr(B) = \frac{9}{10}$$ $$\Pr(C) = \frac{6}{10}$$

What is the probability of at least two events happening?

I'd like to use negation, to negate the possibility that event no event happen plus the probability that only one happens.

$$D = \text{at least two events happen}$$

$$\Pr(D) = 1-\Pr(\text{none happens})-\Pr(\text{exactly one happens})$$

$$\Pr(D) = 1 - \left(\frac{1}{10}\cdot \frac{1}{10}\cdot\frac{4}{10}\right) - \left(\frac{1}{10}\cdot\frac{6}{10}\cdot\frac{1}{10} + \frac{9}{10}\cdot\frac{1}{10}\cdot\frac{4}{10} + \frac{9}{10}\cdot\frac{1}{10}\cdot\frac{4}{10}\right) = 0.918$$

The answer seems a little larger, I can't convince myself that I'm right.

Michael Hardy
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lzc
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2 Answers2

6

Your approach works, and your answer is correct.

It can also be calculated straight forwards as:

$\begin{align} \mathsf P(AB\cup AC\cup BC) & = \mathsf P(AB)+\mathsf P(A^cBC)+\mathsf P(AB^cC) \\[1ex] & = \mathsf P(A)\mathsf P(B) + \bigg(\mathsf P(A^c)\mathsf P(B)+\mathsf P(A)\mathsf P(B^c)\bigg)\mathsf P(C) \\[0ex] &= \frac{9^2}{10^2} + \bigg(\frac{1\cdot 9}{10^2} + \frac{9\cdot 1}{10^2}\bigg)\frac 6{ 10} \\[1ex] & = \frac{918}{1000} \end{align}$

Graham Kemp
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  • why is the first term not $P(ABC^c)$? – Jackson Aug 25 '20 at 19:45
  • We require a union of disjoint events. $$AB\cup AC\cup BC=A(B\cup B^{\small\complement} C)\cup A^{\small\complement} BC$$ An outcome in the union is either in $A$ or not. If it is in $A$, then it is in $B$ or $C$ - which is to say in $B$ xor $B^{\small\complement}C$. Else if not in $A$, then it must be in both $B$ and $C$. – Graham Kemp Aug 25 '20 at 23:27
  • @Jackson In short: We do not want to neglect the outcomes in all three events, $ABC$.$$AB\cup AC\cup BC=ABC\cup ABC^{\small\complement}\cup A B^{\small\complement} C\cup A^{\small\complement} BC$$ – Graham Kemp Aug 25 '20 at 23:37
2

Can confirm that is the right answer, calculated it directly and obtained the same number.

Rammus
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