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 ?
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.
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.