1. ## Probability pdf

Hey all. Here's another I can't get.

The life, X in years of a car part had pdf:

f(x)= 3x² / 7³ * e^-(x/7)³ ; 0 < x < infinity

a) probability that this regulator will last at least 7 years?
b) Given that it lasts 7 yrs, conditional prob. that it will last at least another 3.5 yrs?

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So for a, I got the correct answer by saying p(x>=7) = 1-p(x<=7)
And integrating f(x) from 0--->7

Now for b, would I say 1-integral from 7--> 10.5? Cuz' that's what I was thinking, but it didn't work out. Please explain!

2. So you want $\displaystyle P(X \geq 10.5| X = 7)$. What is the definition of conditional probability?

Originally Posted by amor_vincit_omnia
Hey all. Here's another I can't get.

The life, X in years of a car part had pdf:

f(x)= 3x² / 7³ * e^-(x/7)³ ; 0 < x < infinity

a) probability that this regulator will last at least 7 years?
b) Given that it lasts 7 yrs, conditional prob. that it will last at least another 3.5 yrs?

------

So for a, I got the correct answer by saying p(x>=7) = 1-p(x<=7)
And integrating f(x) from 0--->7

Now for b, would I say 1-integral from 7--> 10.5? Cuz' that's what I was thinking, but it didn't work out. Please explain!

3. The intersection of the two/P(x>=10.5)

How do I get the intersection, the product of the two?
And for p(x>=10.5), do I do 1-p(x<=10.5) from 0-->10.5 ?

4. actually, $\displaystyle P(X \geq 10.5| X = 7) = \frac{P( X \geq 10.5 \ \text{and} \ X = 7)}{P(X = 7)}$. So $\displaystyle X \geq 10.5$ and $\displaystyle X = 7$ are mutually exclusive events. So I think it would be $\displaystyle \frac{P(X \geq 10.5)}{P(X=7)}$. You can't take the product, because they are not independent.

Originally Posted by amor_vincit_omnia
The intersection of the two/P(x>=10.5)

How do I get the intersection, the product of the two?
And for p(x>=10.5), do I do 1-p(x<=10.5) from 0-->10.5 ?

5. Thanks, it worked..
However, I thought that if two events were mutually exclusive, their intersection is 0?

So in that formula, how can you just say to divide the p(x>=10.5/x=7) ?

6. given the context, $\displaystyle X$ is the lifetime of a car. So if $\displaystyle X \geq 10.5$ then it must have reached $\displaystyle X = 7$. So they are not really mutually exclusive. They couldn't be, because then the conditional probability would not make sense.

what was the answer given? was it correct?

7. yea, it was. I thanked" you..lol

I appreciate it