# Bernoulli and Conditional Convergence

• Jul 10th 2010, 07:33 AM
Hitman6267
Bernoulli and Conditional Convergence
The following problem is under the Bernoulli chapter in my book and it's about conditional convergence.
A hospital obtains its flu vaccines:
40% from A
50% from B
10% from C
Amount of ineffective vaccines:
3% in A
2% in B
5% in C

The hospital tests 5 vials from each shipment. If at least one of the five is ineffective, find the conditional probability of that shipment's having come from C.

I know how to solve the problem if the information in bold was not given. But I don't know how the information in bold changes the problem. Can any one explain ? How do I factor in them testing 5 vials.

thank you
• Jul 10th 2010, 08:58 AM
Plato
Quote:

Originally Posted by Hitman6267
The following problem is under the Bernoulli chapter in my book and it's about conditional convergence.
A hospital obtains its flu vaccines:
40% from A
50% from B
10% from C
Amount of ineffective vaccines:
3% in A
2% in B
5% in C
The hospital tests 5 vials from each shipment. If at least one of the five is ineffective, find the conditional probability of that shipment's having come from C.

Suppose that $\displaystyle F$ stands for a batch of five fails.
Then $\displaystyle P(FC)$ is the probability a batch fails from C.
If a batch fails it must come from A, B, or C.
So $\displaystyle P(F)=P(FA)+P(FB)+P(FC)=P(F|A)P(A)+ P(F|B)P(B) + P(F|C)P(C)$.
From the given $\displaystyle P(FC)=P(F|C)P(C)=\left[1-(0.95)^5\right](0.1)$.
You want to find $\displaystyle P(C|F)$.

PS. I have assumed that the hospital tests five vials all from the same supplier.
• Jul 10th 2010, 09:26 AM
Hitman6267
Quote:

Originally Posted by Plato
$\displaystyle P(FC)=P(F|C)P(C)=\left[1-(0.95)^5\right](0.1)$.

My problem lies here. Why is P(F|C) = $\displaystyle 1-(0.95)^5$. For me if we ignore the bolded information P(F|C) is 0.03. How does the bolded information lead to that answer ?
• Jul 10th 2010, 10:03 AM
Plato
Quote:

Originally Posted by Hitman6267
My problem lies here. Why is P(F|C) = $\displaystyle 1-(0.95)^5$. For me if we ignore the bolded information P(F|C) is 0.03. How does the bolded information lead to that answer ?

Where does "P(F|C) is 0.03" come from? It is no where in the given. Is it?
We know that 5% of the vials from C are defected.
We know that 10% of all vials come from C
The test fails if at least on vial is defected: $\displaystyle 1-(.95)^5$.
In other words, not all five pass.

But again what about P(F|C) is 0.03?
• Jul 10th 2010, 10:10 AM
Hitman6267
I think I got it
F|C : event that one of the five vials taken from C is defective.
So P(F|C) = 1 - P(none of the vials are defective)

I mean't 5% instead of 3% in my earlier post.

Thank you
• Jul 10th 2010, 10:25 PM
matheagle
and I so wanted to learn what conditional convergence meant in terms of rvs.