# Probability of being late

I would like help with correcting my solution for the following problem:

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?

Attempted solution

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: $P(A)=0; P(A^c\cap B^c\cap C^c)=0.4; P(C|A\cup B\cup C)=0.6; P(B^c\cap A^c|C)=5/6$

And also:

$P(A^c)=1; P(A^\cup B\cup C)=1-P(A^c\cap B^c\cap C^c)=0.6$ $P(C)=P(C|A\cup B\cup C)*P(A\cup B\cup C)=0.36$ $P(B^c\cap C)=P(B^c|C)P(C)=P(B^c\cap A^c|C)P(C)=5/6*0.36=0.3$ $P(B^c)=P(B^c\cap C)+P(B^c\cap C^c)=0.3+0.4=0.7$

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.

Thanks!

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+1 for showing what you have tried. –  Ross Millikan Dec 3 '11 at 16:18
Sorry, I may be confused, but in the second line of your solution: is $P(C\,|\, [\,A\cup B\cup C\,])=0.6$ right? This is the probability that Clare is late given that at least one of them is late. What you have written is that if at least one is late, then the probability that Clare will be between those who are late is 0.6; which is quite different. This event only occurs when Bob is late and arrives before Carol (I interpret "between" to mean anyone who came immediately before or after Carol was late). Or did you want "between" to be "amongst"? –  David Mitra Dec 3 '11 at 18:41
@David Mitra, Yes, haha, I think my language barrier is showing. "Between those who are late" is to mean that, if someone is late, then the probability of Clare being late is 0.6. (In my native tongue we don't have separate words for 'between' and 'amongst'). Thanks for pointing that out ;). –  josh Dec 6 '11 at 0:01

In a. you found $P(B \cap C^c)=0.24, P(B^c \cap C)=0.3.$ These events are disjoint and are the only ways that exactly one can arrive late, so the chance that Claire arrives late alone is given that only one is late is $\frac{P(B^c \cap C)}{P(B^c \cap C)+P(B \cap C^c)}=\frac{0.3}{0.3+0.24}=\frac{5}{9}$