# Homework Help: Class and student probability question

1. Jan 24, 2010

### Dell

student A comes to 80% of the classes while student B comes to 60%, (A,B are independant of one another), on a given day exactly one student arrived, what is the probability that it was student A

P(A)=0.8
P(B)=0.6

P($$\bar{A}$$)*P(B)=0.12
P($$\bar{B}$$)*P(A)=0.32

C=> only one student arrived
P(C)=P($$\bar{A}$$)*P(B) +P($$\bar{B}$$)*P(A)=0.12 + 0.32 =0.44
P(C/A)=P($$\bar{B}$$)*P(A)=0.32

P(A/C)=$$\frac{P(A)*P(C/A)}{P(C)}$$=$$\frac{0.8*0.32}{0.44}$$=0.58

this is wrong the correct answer is meant to be 0.73, any help??

2. Jan 24, 2010

### Dick

Re: probability

Why isn't it just P(A)*P(not B)/(P(A)*P(not B)+P(B)*P(not A))? Why the extra P(A) factor?

3. Jan 24, 2010

### Dell

Re: probability

apparently it is, but i know P(A/C) is the probability of A if i know that C has occured, and i thought that that was what i needed to use here

4. Jan 24, 2010

### Dick

Re: probability

The numerator is P(A|C). So the whole answer is P(A|C)/P(C). I'm questioning where the extra P(A) came from?

5. Jan 24, 2010

### Dell

Re: probability

as far as i can tell im looking for P(A|C)-> the probability of A if C has occurred,

P(A|C)= P(A)*P(C|A)/P(C)

http://en.wikipedia.org/wiki/Bayes'_theorem

6. Jan 24, 2010

### Dick

Re: probability

Ok, if you want to do it that way, then P(C|A) is the probability that A is there and B is not there divided by probability that A is there. That's 0.32/0.8. You are drawing a Venn type diagram of this, right?

7. Jan 24, 2010

### vela

Staff Emeritus
Re: probability

If you know A arrived, the probability that only one student arrived is the probability that B didn't, so P(C|A)=P(not B).