1
$\begingroup$

I thought about this problem the other day and my limited knowledge in basic probability theory was not enough to help me figure this one out. Say you have 3 probabilities:

  • You have a 50% chance of dying tomorrow due to ant abduction(P(A) = 0.5).
  • You have a 40% chance of dying tomorrow by being beaten by a baboon (P(B) = 0.4).
  • You have a 30% chance of dying tomorrow by being cannibalized by Canadians (P(C) = 0.3).

Then, what is the probability that you will die tomorrow, assuming the previous three reasons are the only ways you could die?

I thought of doing something like P(A or B or C), and since A,B,C are disjoint, then P(A or B or C) = P(A) + P(B) + P(C). This is clearly wrong, since the probabilities can't exceed 1. What would be an appropriate way of thinking about this?

$\endgroup$
  • $\begingroup$ You are correct to notice that $A,B,C$ are not mutually exclusive for the reason that probability cannot exceed 1. This implies that in some of the scenarios of you dying you will be cannibalized by canadians while simultaneously being beaten by baboons and getting abducted by ants. Terrible way to die if you ask me, but makes for an interesting movie idea. In particular, note the problem stated they are independent events. You can prove that if A,B,C are independent, then they are not mutually exclusive. And if they are mutually exclusive, they are not independent. $\endgroup$ – JMoravitz Feb 20 '15 at 21:03
  • $\begingroup$ I see. I initially thought that the events were mutually exclusive, since you can only die once (i.e. if one event happens, the others cannot happen). $\endgroup$ – fparedesg Feb 21 '15 at 1:24
1
$\begingroup$

The probability that you will not die tomorrow, is the probability all events will not happen: $$\mathbb{P}(\text{you will not die})=(1-0.5)(1-0.4)(1-0.3)=\frac{21}{100}$$ So the probability that you will die tomorrow is: $$\mathbb{P}(\text{you will die})=1-\mathbb{P}(\text{you will not die})=1-\frac{21}{100}=\frac{79}{100}$$

$\endgroup$
1
$\begingroup$

what you can do here is instead of looking at your odds of dying, you can look at your probability of living, in this case your probability of surviving A is $.5$, your probability of surviving B is $.6$ and your probability of surviving C is $.7$, so your odds of surviving A and B and C is $.5\times.6\times.7 = .21$ and your probability of dying is $1-P($Surviving$) = .79$

$\endgroup$
-3
$\begingroup$

This problem may be easily solved using the common formula. From the given situation it is obvious that any one of the happenings from three you have described is enough for your death. You tried in good way. Just few more calculations was needed. The formula you used was almost correct. The correct formula is as follows:
$$ \begin{split} P(A\cup B\cup C)&=P(A)+P(B)+P(C)-P(A \cap B)-P(B \cap C)-P(C \cap A) + P(A \cap B \cap C)\\ &= (0.5) + (0.4) + (0.3) - (0.5\cdot 0.4) + (0.4\cdot 0.3) + (0.3\cdot 0.5) + (0.5\cdot0.4\cdot0.3)\\ &= 0.79 \end{split} $$ See more here: https://en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.