Probability Problems 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 ?
 A: 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.
A: 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 $65/324$.
A: 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.
A: 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.
