I would like help with correcting my solution for the following problem:
Adam, Bob and Clare have made an appointment at 5 PM.
Adam is never late.
The probability that none of them is late is 0.4.
If at least one of them is late, the probability Clare will be amongst those who are late is 0.6.
If it is known Clare will be late, the probability she will be the only one who is late is 5/6.
The questions are:
a. What is the probability only Bob will be late?
b. If it is known exactly two people will arrive on time, what is the probability Clare is the one who is late?
I marked $A$ to mean Adam is late, $B$ to mean Bob is late and $C$ to mean Clare.
I got the following information from the question: $P(A)=0; P(A^c\cap B^c\cap C^c)=0.4; P(C|A\cup B\cup C)=0.6; P(B^c\cap A^c|C)=5/6$
$P(A^c)=1; P(A^\cup B\cup C)=1-P(A^c\cap B^c\cap C^c)=0.6$ $P(C)=P(C|A\cup B\cup C)*P(A\cup B\cup C)=0.36$ $P(B^c\cap C)=P(B^c|C)P(C)=P(B^c\cap A^c|C)P(C)=5/6*0.36=0.3$ $P(B^c)=P(B^c\cap C)+P(B^c\cap C^c)=0.3+0.4=0.7$
Using this I calculated a. like this:
a. $P(A^c\cap B\cap C^c)=P(B\cap C^c)=1-P(B^c\cup C)=1-(P(B^c)+P(C)-P(B^c\cap C))=1-(0.7+0.36-0.3)=0.24$
And b. like this:
b. Adam is never late, so he'll always arrive on time and be one of the group which isn't late, so the probability is: $P(C|(A^c\cap B^c)\cup (A^c\cap C^c))=P(C|B^c\cup C^c)=P(C\cap (B^c\cup C^c))/P(B^c\cup C^c)=$ $P(C\cap B^c)/P(B^c\cup C^c)=0.3/(0.7+(1-0.36)-0.4)$
(The last step uses the inclusion-exclusion principle on $P(B^c\cup C^c)$)
I know from my lecturer that at least the answer on b. is incorrect (I'm not sure about a.). But I just can't find my error(s)! I'd really appreciate if someone could point out where I went wrong.