Simple probability question! Having a hard time googling, and I've been away from probability for a bit.
Edit: There's a probability tag here I so figured it was fair game.
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$\begingroup$ P(A or B) = P(A) +P(B) - P(A and B). $\endgroup$– Glen_bCommented Dec 13, 2019 at 6:20
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3 Answers
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Yes --- in fact, this is an axiom of probability
Assuming "dying at X years old" means dying at any time during that year, then yes, that is correct. In fact, this is a direct result of the additivity axiom of probability theory. For any disjoint events $\mathcal{E}_1, \mathcal{E}_2, ..., \mathcal{E}_n$ we have the probability equation:
$$\mathbb{P} ( \mathcal{E}_1 \cup \cdots \cup \mathcal{E}_n ) = \mathbb{P} ( \mathcal{E}_1) + \cdots + \mathbb{P} ( \mathcal{E}_n).$$
(Note: The symbol $\cup$ denotes union of events, which is equivalent to "and".) Other answers have noted that your result can be derived as a consequence of applying the broader probability rule for non-disjoint events, and then recognising that for disjoint events those events can't both occur and so their intersecion is empty. This is really quite a circuitous method, insofar as that rule is derived from the additivity axiom anyway.
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$\begingroup$ It's an axiom in the Kolmogorov formulation. In Cox's formulation, it's derived from the product and sum rules via intermediate results like $P(A \vee B | I) = P(A | I) + P(B | I) - P(A \wedge B | I)$. $\endgroup$ Commented Jan 10, 2020 at 13:34
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$\begingroup$ That's true. Nevertheless, the Kolmogorov axioms seem to have become the standard foundation of probability theory, and the axioms of non-negativity and countable additivity also generally appear as the foundational axioms in broader measure theory. To my mind the Kolmogorov axioms are more directly intuitive and self-evident than the Cox axioms, so I view the standard treatment as better. So I'd say that this is a case where Kolmogorov has probably "won". $\endgroup$– BenCommented Jan 10, 2020 at 13:38
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Yes this is true, one can apply this formula to prove it:
P(A or B)=P(A)+P(B)-P(A and B) which reads : the probability to have A or B equals to the probability to have A plus the probability to have B minus the probability to have both events occurring at the same time (intuitively this last term is to prevent counting a same event twice).
Dying by 65 can be rephrased as dying at 0y.o. or dying at 1 y.o., or dying at 2 yo or ... or dying at 65 yo. Note that here dying at $X$ yo means to die at exactly $X$ yo. It doesn't mean either dying by $X$ nor it doesn't mean dying at $X$ while supposing the person still alive at $X-1$ yo.
Let's P(x) be the probability to die at exactly X y.o.
Then for any x,y, x!=y, P(x and y) =0
(meaning you can't die at two different age, one must pick one's year... )
Then we have:
P(dying by 65) = P(0 or 1 or 2 or.... or 65)
=P(0)+P(1or 2 or... 65) - P(0 and (1 or 2... or 65))
=P(0)+P(1or 2 or... 65) - 0
=P(0)+P(1or 2 or... 65)
= ... = P(0)+ P(1) +... + P(65)
Note: the question can be tricky because it may be understood in different ways:
The probability to be dead by 65 does not equal to the sum over x of the conditional probability to be dead at x y.o. knowing that the person is still alive by X-1 y.o. as another answer illustrated it with an example.
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Let's consider a very dystopian society that deals with overpopulation by combat. At age 20 two enter the arena, only one leaves. This happens again at age 50.
Assuming no other causes of death you clearly have a 25% chance of reaching 65, but simply adding up the odds comes up with a 100% chance you died.
The issue here is that the odds of dying are applied to the results of the previous chances to die. The 50% at age 50 is of the 50% that survived at 20 and thus only 25% of people actually die this way.
