# Conditional Probability

1. Sep 19, 2013

### Woolyabyss

1. The problem statement, all variables and given/known data
Question 1:
A teacher gave his class two tests where every student passed at least one test. 72% of the class passed passed both tests and 80% of the class passed the second test.
(i)what percentage of those who passed the second test also passed the first?
(ii) what percentage of the class passed the first but failed the second test?

Question 2
A factory has three machines P,Q and R, producing large numbers of a certain item Of the total production, 40% is produced on P, 50% on Q and 10% on R. The records show that 1% of the items produced on P are defective, 2% of items produced on Q are defective and 6% of items of items produced on R are defective The occurrence of a defective item is independent of each machine and all other items.
(i)calculate the probability the item chosen is defective.
(ii) Given that the item chosen is defective, find the probability that it was produced on machine
Q.
2. Relevant equations

P(A|B) = P(AnB)/P(B)

3. The attempt at a solution

1. (i)

P(F|S) = .72/.8 = .9

(ii)

Since all students pass at least one test.

P(F U S) = 1

1= P(F) + P(S) - P(FnS).......... P(F) = 0.92 all people who passed the first test

P(F) - P(FnS) = 0.2 This is the number who only passed the first test.

According to the answers in my book I got the first part right but the second part wrong.

Question 2
(i)
PD = .4(.01)
QD = .5(.02)
RD = .1(.06)

(ii)

P(Q|D) = .5(.02)/.5 = .02

According to my book both of these answers are wrong.Any help would be appreciated.

2. Sep 19, 2013

### haruspex

I'd say the book is wrong.
Looks right to me.
Think again about what you are dividing by here.

3. Sep 19, 2013

### Woolyabyss

Would it be P(Q|D) = .5(.02)/.02 = .5 ??

4. Sep 20, 2013

### haruspex

Yes. Do you see how that follows from the equation?

5. Sep 20, 2013

### Woolyabyss

Ya, I didnt divide by the probability of q being defective the first time. Thanks