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The exercise

A focus consists of $20$ bulbs that operate independently.

The lifetime of each bulb is a continuous random variable $X$, whose density function is as follows:

$$\displaystyle F_X(x)=1-e^{-\dfrac{x}{50}}$$$$x>0$$

When more than $2$ light bulbs are failed, the maintenance service is called to renew the focus.

What is the probability that in $10$ days the maintenance service is not required?

My solution

-I will use in the distribution of a discrete random variable, the probability of a continuous random variable-

Let Y be the number of bulbs that do not function when analyzing $n=20$ of them, each with an independent probability $p=1-F_X(x)$ to fail, that is: not having $x$ days of useful life. Then, $Y$~$B(n,p)$ so that:

$$\displaystyle f_Y(x,y)=C_y^{20}(1-F_X(x))^y(F_X(x))^{20-y}$$$$x>0$$$$y=0,...,20$$

And if all of the above was right, the probability would be:

$$1-P(Y>2)=P(Y≤2)=f_Y(10,0)+f_Y(10,1)+f_Y(10,2)$$

Is it okay to do what I did? The solution is correct? Thank you very much.

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1 Answer 1

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Yes, indeed, your approach is okay. However, do notice that your $f(x,y)$ is the probability that exactly $y$ of the $20$ bulbs will expire after $x$ days.

$$\begin{align} f_Y(x,y) &= \mathrm C_y^{20} \,\mathsf P(X>x)^y\,\mathsf P(X\leq x)^{20-y} \\[1ex] & = \mathrm C_y^{20}\big(1-F_X(x)\big)^y\big(F_X(x)\big)^{20-y}\end{align}$$

For the maintanence not to be required you need less than three of the twenty bulbs to expire within the ten day period.   That is that no less than 19 survive.

That is, you seek: $f(10,20)+f(10,19)+f(10,18)$

That is all.

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  • $\begingroup$ But $Y$ represents the number of failed bulbs. When $(Y> 2)$ then maintenance is called. But we want to know what the probability is that they do not need the maintenance, so it would be $1-P(Y>2)=P(Y≤2)=f_Y(10,0)+f_Y(10,1)+f_Y(10,2)$. I am right? $\endgroup$ Commented Feb 15, 2017 at 4:29
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    $\begingroup$ $\mathsf P(X>x)^y$ is the probability that a set of $y$ bulbs all survive $x$ days without failure. Your function $f_Y(x,y)$ does not measure what you intended. $\endgroup$ Commented Feb 15, 2017 at 4:40

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