Probability question about exponential distribution

[dch]

Homework Statement

A manufacturing process produces 92% good chips (G) and 8% bad chips (B).
The lifetime, in seconds, of chips is exponentially distributed [tex]E(\lambda)[/tex].
For good chips, [tex]\lambda1=1/20000[/tex] For bad chips,[tex]\lambda2=1/1000[/tex]
Every chip is tested for 50 seconds prior to leaving the factory. The only chips that pass the test will be sent out to customers (S).

Homework Equations

a. Find the probability that a good chip will be sent out to customers.
Find the probability that a bad chip will be sent out to customers.
b. Evaluate the percentage of good chips among a great lot of chips that will be sent out to customers.

The Attempt at a Solution

a.
I used this equation to solve :
[tex]P(X>50)=1-P(X<=50)=1-(1-e^{-\lambda*50})[/tex]
So P1(X>50)=0.9975
P2(X>50)= 0.9512

Am I right? Question b I don't know how to solve please suggest me asap. Tomorrow I will have the final test.

Many thanks.

 

[grey_earl]

Seems good to me, although I'm not an expert in probability theory. For the second question, you know that 0.08 of the produced chips are bad, they are sent out to customers with a probability of 0.9512, so 0.0761 of produced chips are bad and sent out, while 0.92*0.9975 = 0.9177 of produced chips are good and set out. That makes the percentage of bad chips of the ones which are sent out 0.0761/(0.0761+0.9177) = 7.66% which is just slightly below 8%. Not very effective, this test process.

 

[rn315656]

[tex]\int \sqrt{\frac{x^4-1}{x^6}}dx=\int \sqrt{\frac{1}{x^2}-\frac{1}{x^6}}dx[/tex]

 

[dch]

Thanks grey_earl, so conclusion:

100 chips are manufactured in the beginning. Of these, 92 are good and 8 are bad. 99.75% of the 92 good chips pass the test = 91.77 good chips. 95.12% of the 8 bad chips pass the test = 7.61 bad chips. Altogether, 91.77 + 7.61 = 99.38 chips pass the test. The fraction of these that are good is 91.77 / 99.38 = 0.9234.

 

[grey_earl]

Yes!

 

[Ray Vickson]

Say "OK" occurs if the chip is sent out. We have P{OK|good} = 0.9975 and P{OK|bad} = 0.9512 (as in your computations). You want to know P{good|OK}. What quantities do you need to know to get this? Hint: Bayes' formula and the definition of conditional probability.

RGV