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Problem:

The probability that a man who is 85 years. old will die before attaining the age of 90 is $\frac13$. A,B,C,D are four person who are 85 years old. what is the probability that A will die before attaining the age of 90 and will be the first to die ?

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  • $\begingroup$ @Alex: Umm I am not quite sure .. but does the first answer is $\frac{1}{5}\times \left[1+1+0+\frac{1}{2} +\frac{1}{2}\right] = \frac{3}{5} $??! I don't have the solution so I can't check. $\endgroup$ – Quixotic Dec 9 '10 at 13:54
  • $\begingroup$ @ Alex:what I did is the just used the fact that the event are mutually exclusive and hence ...But I am a bit interested in what do you meant by precisely tree of probabilities? $\endgroup$ – Quixotic Dec 9 '10 at 13:59
  • $\begingroup$ @Alex:Where is the answer that you posted?I am really much curious to know how would your approach leads to the correct answer. $\endgroup$ – Quixotic Dec 10 '10 at 13:26
  • $\begingroup$ I reposted my answer in such a way that it requires close to zero thinking. Please kindly let me know, where the error lies. $\endgroup$ – Alex B. Dec 10 '10 at 14:32
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    $\begingroup$ Just to explain to uninitiated bystanders: I tried to guide Debanjan to the answer with hints. Since he wasn't able to finish by subtracting two numbers (that he had already computed) he told me that my method was wrong, as confirmed by an IMO gold medalist friend of his, and that he now got the answer from this friend. I therefore deleted the old post, since the long thread of hints was evidently wasted on Debanjan. $\endgroup$ – Alex B. Dec 10 '10 at 14:35
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The probability that A dies first is 1/4 (since it's the same probability for everyone).

The probability that they all die after the age of 90 is $(2/3)^4 = 16/81$. The two events are independent, so the probability that A dies first and they all die after the age of 90, or equivalently that A dies first and that that happens after he is 90, is $\frac{16}{81}\cdot\frac{1}{4}=\frac{4}{81}$. But if A dies first, then either he dies first and is above 90 or he dies first and is younger than 90, and the two are mutually exclusive. Therefore (or equivalently using the rule that $P(X) = P(X\cap Y) + P(X\cap Y^C)$), we deduce that the probability that A dies first and before the age of 90 is 1/4 - 4/81 = 65/324.

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  • $\begingroup$ +1 I like this answer, not only does it confirm mine, of which I was unsure, but it is much simpler. $\endgroup$ – yunone Dec 10 '10 at 14:42
  • $\begingroup$ @yunone Thanks. This was my first answer (but only with hints). I deleted it because it was of no use to the OP. $\endgroup$ – Alex B. Dec 10 '10 at 14:44
  • $\begingroup$ +1,I learned probablity just couple of days back ... anyways,it was simple sorry again for suspecting you :) But am sure you are aware that I have no intention of offense. $\endgroup$ – Quixotic Dec 10 '10 at 14:51
  • $\begingroup$ I really feel like banging my head ... it was so simple ... damn! $\endgroup$ – Quixotic Dec 10 '10 at 14:54
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This is essentially the same approach as Alex Bartel's, expressed perhaps more simply.

The probability that at least one dies by age 90 is $1 - (2/3)^4 = 65/81$, and when that happens the probability that it's A who dies first is (by symmetry) a quarter of that, or $65/324$.

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I'm not sure from the comments above if you've solved this or not already, but I had this idea when I first read your post.

I think you can partition the outcome space, based on when $B,C,D$ die. There are four disjoint cases:

$B_0:$ none of them dies before 90,

$B_1:$ exactly one of them dies before 90,

$B_2:$ exactly two...

$B_3:$ exactly three...

So $$P(B_n)=\binom{3}{n}(1/3)^n(2/3)^{3-n}.$$

Then let $F$ be the event that $A$ dies first, and $E$ be the event that it happens before $90$.

So $$P(F\cap E)=\sum_{n=0}^3 P((F\cap E)|B_n)P(B_n)$$

So for example, $P((F\cap E)|B_0)$ is the probability that $A$ dies first and before $90$, given that the other three die after $90$. This occurs with probability $(1/3)$, since all the others are given to die after $90$. Then $P((F\cap E)|B_1)$ is the probability that $A$ dies first and before $90$, when exactly one other person dies before $90$. This probability is $(1/3)(1/2)$. The other cases are similar, and then one can just add them up.

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The probability of dying by 90 is the same for all four. The probability that A dies first is simply 1/4 since we are give no more information. Since the two events are independent, the probability of their conjunction, i.e., that A dies and is the first to die, is simply the product of the two probabilities, or 1/12.

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    $\begingroup$ This is not correct, as the previous answers have shown. You have missed that if A dies first, it is more likely than 1/3 it was before age 90, so the events are not independent. $\endgroup$ – Ross Millikan Mar 15 '11 at 13:55

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