Question

A hospital receives 2/5 of its flu vaccine from Company A and remainder from Company B. Each shipment contains a large number of vials of vaccine. From Company A, 3% of the vials are ineffective; from Company B, 2% are ineffective. A hospital tests n =25 randomly selected vials from one shipment and finds that 2 are ineffective. What is the conditional probability that this shipment came from Company A?
Please, help. It's so complicated to me.

Answer 1

P(A) *P(2 ineffective | A) = 2/5 * 25c2*0.03^2*0.97^23 = 2/5 *0.134002731872 = X
P(B) *P(2 ineffective | B) = 3/5 *25c2*0.02^2*0.98^23 = 3/5 *0.075401673858 = Y

P[conditional probability it came from A] = X /(X+Y) = 0.54229

Answer 2

^This appears to be correct, but maybe I'll elaborate to make notation easier. There is no defined notation up there

Let \(S\)= shipment from company A
\(I\) = ineffective vial

You are given:
\(P(S)=2/5 = 0.4\)
\(P(I|S) \) This is essentially a binomial probability , you are given the success probability to be 0.03, so, and there are essentially a total of 25 trials with 2 successes.
\[ P(I|S)={ 25\choose 2}(0.03)^2(1-0.03)^{23}\]

\[P(I|\overline{S})={ 25 \choose 2}(0.02)^2(1-0.02)^{23}\]

You are asked to find \(P(S|I)\)

By Bayes' theorem:

\[ P(S|I)=\frac{P(I|S)P(S)}{P(I|S)P(S)+P(I|\overline{S})P(\overline{S})}\]

Answer 3

I don't get why \(P(I|S)={ 25\choose 2}(0.03)^2(1-0.03)^{23}\)
Please, explain more

Answer 4

Let the combination of the situation is 2 IE + 23 E, the possibility that combination happen is \(\dfrac{25!}{2!*23!}\) I got this part. but not the second term and the third term

Answer 5

So P(I|S) is the probability of the vials being ineffective, given it's from shipment of company A. So, they say that 3% of the vials are ineffective. You can think of this as being the "success" probability. They find out that out of 25 vials, 2 are ineffective. So, you can think of them are doing 25 independent Bernoulli trials, where the probability of an outcome is the same for every trial, where there are 2 "successes" . Or if you think of it this way maybe, they know 3% of the vials are ineffective. If 25 of them are selected at random (i.e. independently), what is the probability that 2 vials are ineffective? This is basically the binomial distribution.

Answer 6

Thanks for the help. I need time to digest it. :)

Answer 7

If you understand the binomial coefficient, then understanding the other 2 probabilities shouldn''t be too difficult. Consider one sequences in which you get 2 successes, and then 23 failures in a row (i,e, 2 ineffective vials, and 23 okay vials))

Then you have:
1st trial is a success, with probability 0.03
2nd trial is a success with probability 0.03
So so far, you have (0.03)(0.03)=(0.03)^2

Then, 3rd trial is a failure, with probability 1-0.03 = 0.97
So you have (0.03)^2*(0.97)

Continue like this for 25 trials in total, and you get
\[(0.03)^2(0.97)^{23} \]

Then you have the binomial coefficient to determine all combinations possible of rearrange 2 successes and 23 failures

Answer 8

nvm, I got your logic. hihiihi.

Answer 9

One more question: why do we pick the third term of the polynomial to calculate?
I mean \((0.03+0.97)^{25}\) And after expand them we pick \(25C2) (0.03)^2 (0.97)^23\)

Answer 10

Because the third term represent the first 2 vials are ineffective and the third one is effective, am I right?

Answer 11

I wouldn't necessarily think of this probability problem as coming from a binomial expansion, because really what you have is just \((1)^{25}\). But let's look at a simpler example and then generalize to your problem.

Say you roll a die (6-sided die). Say you roll it 3 times. What is the probability of rolling a "2" on the 1st roll only? This is 1/6 * (5/6) * (5/6), since the 2nd and 3rd rolls cannot be a "2" Now, what is the probability of rolling one "2" only? (Like a "2" only once for all 3 rolls), this is a binomial probability in the same way ... We may obtain a "2" on the 1st rool, or 2nd roll, or 3rd roll, so we have \[ {3 \choose 1}(1/6)(5/6)^2\]

It's not necessarily the order you roll it in, but there is only an exponent 1 on 1/6 since you only get ONE success probability, and then you get TWO failure probabilities (for the (5/6) ).

Then just extrapolate this thinking to our problem, except that the probability of success is 0.03, and of failure its 0.97, and there are 2 successes and 23 failures.

Answer 12

hey, it is not just (1)^25, that is (0.03 IE + 0.97 Effective)^25, how can you combine them to get 1^25? hihihi

Answer 13

Well it's just looking at the numbers themselves lol. But it is 1^{25} if you binomially expanded all 26 terms , it will all add to 1 :)

Answer 14

I'm wondering what classes are you taking? You seem to have questions from all different subjects hehe

Answer 15

OOOK, I got you now. Again, thank you.

Answer 16

I have 3 courses: abstract algebra (it is about permutation, polynomials, group theory)
advance calculus ( it is about limits, theoretical linear, fields)
and probability and statistics ( this problem)
hihihi and I got mad with math. All are proving and proving.

Answer 17

Those classes are arranged in Fall only. So that, if I don't take them simultaneously, I have to spend 1more year to finish the compulsory courses.

Answer 18

ah ok, well good luck ;)