the probability that a man who is 85 yrs old will die before attaining the age of 90 is 1/3. Four persons A1,A2,A3, and A2 are 85 yrs old. The probability that A1 will die before attaining the age 90 and will be the first one to die is
$\begingroup$
$\endgroup$
10
-
5$\begingroup$ This is a less routine question than most basic probability questions. It does not yield immediately to mechanical manipulation. $\endgroup$– André NicolasCommented Apr 29, 2013 at 13:24
-
5$\begingroup$ @user1729: The question (apart from being more interesting than many) seems no more or less quoted than many others. If there were clear criteria for closure in such cases, and they were applied fairly systematically, then I would understand (though perhaps not approve). $\endgroup$– André NicolasCommented Apr 29, 2013 at 15:12
-
4$\begingroup$ @user1729: I would understand why the closure. But when this is done very sporadically, when it is done it appears arbitrary. $\endgroup$– André NicolasCommented Apr 29, 2013 at 15:31
-
4$\begingroup$ This question was just reopened and is now about to be closed again. Which is retarded. $\endgroup$– JimCommented Apr 29, 2013 at 17:57
-
4$\begingroup$ @Noah: No. It doesn’t match the standards of some MSE users. As anyone reading the comments can see, other users disagree. $\endgroup$– Brian M. ScottCommented Apr 30, 2013 at 6:04
|
Show 5 more comments
2 Answers
$\begingroup$
$\endgroup$
1
Assume independence. Beside A_1, there could be $0$, $1$, $2$, or $3$ people who die before age $90$.
The probability that A1 dies and the others don't is $\left(\frac{1}{3}\right)\left(\frac{2}{3}\right)^3$. If this happens, then A1 is sure to be first.
The probability that A1 and exactly one other person dies before $90$ is $\binom{3}{1}\left(\frac{1}{3}\right)^2\left(\frac{2}{3}\right)^2$. If this happens, then A1 is first with probability $\frac{1}{2}$.
The probability that A1 and exactly two other people die before $90$ is $\binom{3}{2}\left(\frac{1}{3}\right)^3\left(\frac{2}{3}\right)$. If this happens, then A1 is first with probability $\frac{1}{3}$.
The probability that everybody dies before $90$ is $\left(\frac{1}{3}\right)^4$. If this happens, then A1 is first with probability $\frac{1}{4}$.
Add the numbers obtained for the $4$ cases to find the probability.
answered Apr 29, 2013 at 17:07
-
$\begingroup$ The way I read the question, "first to die" simply happens in 3! cases (the orders in which the other three people die) out of 4! cases (all the possible arrangements of the orders of death), which means that the person has a 1/4 chance of being the first to die, regardless of the age of anyone when they die. Of course this is also what we expect if we select the order of deaths by drawing names out of a hat. $\endgroup$ Commented May 1, 2013 at 13:27
$\begingroup$
$\endgroup$
A1 can be the first to die: if dies and is the only one who does, probability $=\frac 13\cdot (\frac 23)^3$, or if he is one of two who die and he is first, probability $(\frac 13)^2(\frac 23)^2 \frac 12$, or (just keep going)
answered Apr 29, 2013 at 16:54