Question

Suppose that components have failure times that are independent and that can be modeled
according to an exponential distribution with an expected value of 100 days. If a
box contains 5 components, what is the probability that the box has at least 3 components
that last longer than 75 days?

Answer 1

The main problem asks if related to a binomial distribution. You have that if \(X\) is the number of components lasting longer than 75 days, then \(X\sim Binomial(5, p)\). Since they want at least 3 components, you should find \(P(X \le 3)\).
But the success probability \(p\) is unknown, but this would be the probability of the component lasting longer than 75 days. This is actually a random variable itself, say \(Y\) and \(Y\sim Exponential~distribution\). Depending on the parameterization you use for the exponential, since it has a mean of 100, i.e. \(E(Y)=100\), then you can find the parameter value of your exponential based on that. So, \(p=P(Y>75)\)

Answer 2

oh sorry at least 3 components should be \(P(X \ge 3)\)

Answer 3

ok so I what I have currently is that P(X <= 75) = 1-e^(75/100) and I think the P(X>75) would be 1-P(X<=75)

Answer 4

I'm not completely sure how I should apply this to the Binomial distribution to get the at least 3 out of 5 part

Answer 5

oops the exponential distribution one should be 1-e^(-75/100)

Answer 6

Ya the exponential part looks right. For the other part... I'm not sure if there is a formula for the binomial distribution (or if there is, it is probably a bit complicated),

Answer 7

but since it is a discrete distribution, you can find simply that \(P(X \ge 3)=P(X=3)+P(X=4)+P(X=5)\) (since there are 5 "trials" (components chosen) in total)

Answer 8

formula for cdf of the binomial distribution*

Answer 9

I was thinking that too, but I was wondering if I was wrong since there could be cases where it is at last 500 out of 1000 then you couldn't just add them.

so P(X>=3)=(P(X > 75))^3+(P(X > 75))^4+(P(X > 75))^5

Answer 10

I think?

Answer 11

I would say that the binomial distribution has the following pmf:
\[P(X=x)={n \choose x}p^x(1-p)^{n-x}, ~~x=0,1\ldots, n\] , where \(n=5\). But we don't know \(p\), but this success proability is the probability of the component lasting longer than 75 days, which could be found as \(P(Y>75)\) . Once you find that, that value is \(p\) in the binomial distribution. Since you want the probability of 3 out of 5 components lasting longer than 75 days, you'd say \[ P(X \ge 3)=P(X=3)+P(X=4)+P(X=5)\\
={5 \choose 3}(P(Y>75))^3(1-(P(Y>75))^{5-3}
\\+{5 \choose 4}(P(Y>75))^4(1-(P(Y>75))^{5-4}\\
+ {5 \choose 5}(P(Y>75))^5(1-(P(Y>75))^{5-5} \]
It looks a bit weird with the \(P(Y>75)\) plugged into the binomial liked that, but if you calculate it first, will will just be a number to plug in :)

Answer 12

ok, that makes sense to me. That binomial theorem matches the one from my notes. Thanks a lot for your help!

Answer 13

awesome =]