Probability of Low Grade Gas Shipment from Two Plants

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I've been working on a problem and was wondering if someone could check and see if I am on the right track.

A company produces gas from two plants, A and B. (both are considered to be continuous randm variables; X and Y respectively)

For Plant A, its probability density function is:

f(x) = 0.005(x-80) for 80<x<100
0 otherwise

For plant B, it's probability denisity function is:

f(y) = 0.02(y-80) for 80<y<90

r is the octane rating of the gasoline and the gas is:

low grade if r<85
High grade if r>=85

There is an equal probability that the gas was produced at plant A or plant B.

a. What is the probabilty that today's shipment is low grade?
b. If it is low grade what is the probabiliy that it came from plant A?

Here are my answers:

a. f(x) = .0615 (for r<85)
f(y) = .25 (r<85)
The probability that it is low grade is f(x) +f(y) = .3125

b. P(A|(r<85))

P(A) = 1/2
P(r<85) = .3125

P(A|r<85) = P(A and r<85)/P(r<85) = I don't know if this is the right set up for this portion of the problem

Thanks for the help!
 
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To find low grade probability aren't you supposed to be integrating f(x) and f(y) over 0 < r < 85?
 
I am sorry I did not note that I performed the necessary integration for f(x) and f(y) to determine their respective values.
 
So you mistyped F(.) as f(.)? E.g. Fx(85) = \int_0^{85}f_x(x)dx = .0615?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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