# How to find the probability in this continous case?

The lifetime in hours of a certain kind of radio tube is a random variable having probability density function

$$f(x) = \begin{cases} \frac{100}{x^2} , & x>100 \\ 0, &otherwise\\ \end{cases}$$

What is the probability that exactly 2 of 5 such tubes in a radio set will have to be replaced within the first 150 hours of operation?

MY ATTEMPT:

To be quite honest with you I'm having trouble understanding this question

But I tried to use:

$$p(X>100; n=0.4) = \int_{100}^{150} f(x) * (0.4)$$

I'm pretty sure this is wrong but I'm not entirely sure of another way that this can be done.

This is not homework. I'm studying past-exams for an exam I have on Tuesday.

Some help would be great.

-

Your integral will give you the probability that a single radio tube will break in the first $150$ hours. So for each radio tube $$\Pr(\text{break}) = \int_{0}^{150} f(x)\ dx = \int_{100}^{150}\frac{100}{x^2}\ dx = -\frac{100}{x}\bigg|_{100}^{150}=\frac{1}{3}$$ This is the probability of a single tube breaking in the first $150$ hours. Now it's safe to assume that the breaks are independent of each other. We can therefore rephrase your question as
Given that each radio tube breaks with probability $\frac{1}{3}$ in the first $150$ hours, what is the probability that exactly $2$ of $5$ tubes break?
This is given by a binomial probability distribute. We have $2$ "failures" (the breaks) which occur with probability $p = \frac{1}{3}$ and we have $3$ successes with probablity $1-p = \frac{2}{3}$. Therefore the net probability is $$\Pr(2\ \text{of}\ 5\ \text{breaks})=\binom{5}{2}\left(\frac{1}{3}\right)^2\left(\frac{2}{3}\right)^3 = \frac{80}{243} \approx 0.329$$
Find the probability of one of the tubes being replaced in the first $150$ hours, then use a binomial random variable to find the probability that exactly $2$ out of $5$ of them will have to be replaced in the first $150$ hours.