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I have a question which I haven't been able to figure out.

Given three disjoint events $X,Y,$ and $Z$ find

a) the probability that exactly one event occurs

b) the probability that exactly two of the events occur.

So far, the progress I have made on a) is $$P(\text{exactly one}) = [P(X)(1 - P(Y))(1 - P(Z))] + [P(Y)(1 - P(X))(1 - P(Z))] + [P(Z)(1 - P(X))(1 - P(Y))].$$

I took a similar approach for the second one, too. However, given that these sets are disjoint, I think that I might be headed in the wrong direction.

questionasker
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askdokTheBest
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    If by disjoint you mean mutually exclusive then knowing that, say, $X$ occurs already tells you that $Y,Z$ do not occur. So the answer to part $a$ is $P(X)+P(Y)+P(Z)$ and the answer to part $b$ is $0$. Otherwise, what did you mean by "disjoint"? – lulu Feb 04 '21 at 23:05
  • Indeed, the definition of disjoint is the key to this question. Disjoint events have no common outcomes; they are mutually exclusive; the cannot occur simultaneously; et cetera. – Graham Kemp Feb 04 '21 at 23:07

1 Answers1

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(a) The probability you are interested in is given by: \begin{align*} \mathbb{P}(A\cap B^{c}\cap C^{c}) + \mathbb{P}(A^{c}\cap B\cap C^{c}) + \mathbb{P}(A^{c}\cap B^{c}\cap C) \end{align*}

Since $A\cap B = \varnothing$, we conclude that $A\subseteq B^{c}$.

Moreover, since $A\cap C = \varnothing$, we conclude that $A\subseteq C^{c}$.

Therefore $A\subseteq B^{c}\cap C^{c}$, whence we get that $A\cap B^{c}\cap C^{c} = A$.

Similarly, $A^{c}\cap B\cap C^{c} = B$ and $A^{c}\cap B^{c}\cap C = C$, and we are done.

(b) As @lulu has mentioned, such probability equals zero because $A$, $B$ and $C$ are pairwise disjoint.

More precisely, one has \begin{align*} \mathbb{P}(A\cap B\cap C^{c}) + \mathbb{P}(A\cap B^{c}\cap C) + \mathbb{P}(A^{c}\cap B\cap C) = 3\times \mathbb{P}(\varnothing) = 0 \end{align*}

Hopefully this helps!

user0102
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  • We have not been provided that the events are exhaustive, merely that they are mutually exclusive (aka disjoint). – Graham Kemp Feb 05 '21 at 01:23
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    @GrahamKemp Thanks for pointing that out. I have fixed my argument. I think it is ok this time. – user0102 Feb 05 '21 at 01:41
  • Thanks for the hint. For the sake of simplicity -> what would be the difference if the events are mutually exclusive? – askdokTheBest Feb 07 '21 at 20:05
  • Maybe I have not understood your question, but we have assumed they are mutually exclusive in order to solve the problem. – user0102 Feb 07 '21 at 20:29
  • I had worded my last comment incorrectly, so let me rephrase: For the sake of curiosity(and understanding), how would it change given that these events are mutually exclusive? – askdokTheBest Feb 07 '21 at 21:43
  • If I got it right, the fact that they are mutually exclusive implies that

    $$\mathbb{P}(A\cap B^{c}\cap C^{c}) = \mathbb{P}(A)$$

    Similar results apply to the other two cases. Is this what you are looking for?

     – user0102 Feb 08 '21 at 00:07
  • @askdokTheBest Disjoint universally implies mutually exclusive; the converse holds when the sample space is finite. $\quad$ There are conflicting definitions on whether mutual exclusivity (correspondingly: disjointedness) is defined collectionwise or pairwise. This answer, which I agree with, uses the latter definition. $\quad$ If it is given only that the events are mutually exclusive, then this answer can be tweaked to give the same results, as suggested by user0102's comment above this. – ryang Mar 04 '23 at 20:34