Question

I need help in R, on how to graph the distribution to the answer to the problem below and screenshot the work please. Thank you.

Assume that a component passes a test is 0.85 and that components perform independently. What is the probability that the third failure will occur on the tenth component tested? Probability that component passes a test is 0.85 then, probability that failure of component = 1 -0.85 = 0.15 Required probability = (probability that two component will exactly fail among the first 9 component tested) X(probability of failure of 10 component) =9 X 8 * (0.15)?(0.85)? 1 X 2 = 0.2597 Probability of tenth component tested is 0.15 because components performed independently so probability of failure for each component remain the same 0.15 Therefore, the third failure will occur on the tenth component tested is = 0.2597 * 0.15 = 0.03895

Answer 1

Let X be a random variable denoting the number of trials until the third failure.
It is given that P(Failure) = 0.15.

Modeling of the p.m.f.
The last trial results in the third failure. That is in the
th
trial, the third failure occurs.
Out of the remaining (x-1) trials, 2 failures have occured.
The probability of 2 failures is $\large 0.15^2$ for these two failures.
Out of the first (x-1) trials, these two succeses can occur in
C -1 2
ways.
The probability of (x-3) successes is $\large 0.85^{x-3}$

Hence,
$\large P(X=x)=\binom{x-1}{2}0.85^{x-3}*0.15^2*0.15$
$\large \Rightarrow P(X=x)=\binom{x-1}{2}0.85^{x-3}*0.15^3$

Clearly, $\large x\geq 3$

This is coded in R as:
> x=seq(3,30,1)
> prob=choose(x-1,2)*0.85^(x-3)*0.15^3
>
> plot(x,prob,type="h",col=ifelse(x==10, "red", "black"))

We get the following graph:

prob 0.01 0.02 0.03 0.04 on 10 15 х 20 25 30


I hope this clarifies your doubt. If you're satisfied with the solution, hit the Like button. For further clarification, comment below. Thank You. :)