Bernoulli and Binomial Distribution

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dawsob3

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A hospital obtains 40% of its flu vaccine from company A, 50% from Company B and 10% from Company C. From past experience it is known that 3% of the vials from A are ineffective, 2% from B are ineffective, and 5% from C are ineffective. The hospital tests five vials from each shipment, if at least one of the 5 is effective, find the conditional probability of that shipment having come from Company C.

I believe that the answer is 0.1780
Hint: You need to find the probablility of at least one ineffective if from Company A, B or C (1-P(0 defective))

I have been at this for 3 hours and I CAN NOT come up with a solution. It would be so much appreciated if someone could fully explain step by step for me. THANK YOU SO MUCH.

I know that Binomial Distribution is X bin (n,p) X=x when (nCx) (p)^x (q)^n-x
I believe n=5. I have no idea what p, q, and x are. I tried different numbers for each, like i said for 3 hours and can't come up with an answer to match. Maybe this answer isn't even correct? PLEASE i need help! I just want to cry I'm so frustrated.

Once I get the denominator, which is the sum of each different company, I take C and divide it by that. But, I can't get there.
 
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dawsob3 said:
A hospital obtains 40% of its flu vaccine from company A, 50% from Company B and 10% from Company C. From past experience it is known that 3% of the vials from A are ineffective, 2% from B are ineffective, and 5% from C are ineffective. The hospital tests five vials from each shipment, if at least one of the 5 is effective, find the conditional probability of that shipment having come from Company C.
Click to expand...
I will help you set this up.
Let \(\displaystyle F\) be the event that at least one vial fails.
You want to find \(\displaystyle \mathcal{P}(C|F)\).
That is
\(\displaystyle \mathcal{P}(C|F)=\dfrac{\mathcal{P}(F|C)\mathcal{P}(C)}{\mathcal{P}(F|A)\mathcal{P}(A)+\mathcal{P}(F|B)\mathcal{P}(B)+\mathcal{P}(F|C)\mathcal{P}(C)}\)

Here is a hint: \(\displaystyle \mathcal{P}(F|C)\mathcal{P}(C)=(1-(.95)^5)(.10)\)
 
pka said:
I will help you set this up.
Let \(\displaystyle F\) be the event that at least one vial fails.
You want to find \(\displaystyle \mathcal{P}(C|F)\).
That is
\(\displaystyle \mathcal{P}(C|F)=\dfrac{\mathcal{P}(F|C)\mathcal{P}(C)}{\mathcal{P}(F|A)\mathcal{P}(A)+\mathcal{P}(F|B)\mathcal{P}(B)+\mathcal{P}(F|C)\mathcal{P}(C)}\)

Here is a hint: \(\displaystyle \mathcal{P}(F|C)\mathcal{P}(C)=(1-(.95)^5)(.10)\)
Click to expand...

I did this exact approach. Hm, maybe my computing is wrong. Could you maybe explain as to why we are choosing to do... 1-.95. instead of 1-.05

I think that part is confusing me. Also, why exactly we are raising it to the 5th power?

The formula that I posted, does that not apply here to this problem? Binomial Distribution? Is this not that. Maybe that's why i am lost!

I did this whole approach and did not end up with : 0.1780
I found this solution on a practice test. Maybe that is wrong? I'm so confused. This question was in the Bernoulli and Binomial Section of the Text

Thanks so Much
Brittany
 
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dawsob3 said:
I did this exact approach. Hm, maybe my computing is wrong. Could you maybe explain as to why we are choosing to do... 1-.95. instead of 1-.05
Click to expand...
If we test five vials from company C the probability that none of those is defected is \(\displaystyle (.95)^5\).
So the probability that at least one is defective is \(\displaystyle 1-(.95)^5\).
 
pka said:
If we test five vials from company C the probability that none of those is defected is \(\displaystyle (.95)^5\).
So the probability that at least one is defective is \(\displaystyle 1-(.95)^5\).
Click to expand...

Hm, I'm still confused. Because the question is asking for at least one is effective. if I do 1- effective... isn't that now turning it into INEFFECTIVE. (the compliment?)

Could you quickly explain if this is Bernoulli? I am not seeing how it relates to this theory.

thanks again so much!
Brittany
 
dawsob3 said:
Because the question is asking for at least one is effective. if I do 1- effective... isn't that now turning it into INEFFECTIVE. (the compliment?)
Could you quickly explain if this is Bernoulli? I am not seeing how it relates to this theory.
Click to expand...
First it is indeed Bernoulli. Butwe don't really use it.

Test one vial from supplier C.
The probability it is INEFFECTIVE equals \(\displaystyle 0.05\).
The probability it is EFFECTIVE equals \(\displaystyle 0.95\).

Test five vials from supplier C.
The probability all are INEFFECTIVE equals \(\displaystyle (0.05)^5\).
The probability all are EFFECTIVE equals \(\displaystyle (0.95)^5\).

The probability that at least one of the five is INEFFECTIVE is one minus the probability that all are EFFECTIVE.
 
pka said:
The probability that at least one of the five is INEFFECTIVE is one minus the probability that all are EFFECTIVE.
Click to expand...

ok ok, i get those parts now! thanks for much.

this is still confusing me though! haha. I am sitting here trying to figure it out. so,

"The hospital tests five vials from each shipment, if at least one of the 5 is effective, find the conditional probability of that shipment having come from Company C."

see, the question is stating, if at least one of the 5 is EFFECTIVE. then, wouldn't it be 1-(.05)^5?
this would be... the probability that at least one of the five is EFFECTIVE is one minus the probability that all are INEFFECTIVE.

omgoodness, am i just being dumb?
im so sorry... i seriously can not understand this!

also, so if i were to address this problem (we have to fully explain in a report... how would i bring up bernoulli? thanks so so so much!)
 
The example I gave you is for at least one is INEFFECTIVE.
You can do it for for at least one is EFFECTIVE.
 
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