The first principles here are the definitions of conditional probability and the definition of logical connectives.
That is, start with
P(X|Y) = P(XY)/ P(Y)
We know A^A <==> A
Let X = A and Y = B^A
therefore, P(A|B^A) = P(A^(B^A) )/ P(B^A) = P (B^A) /P(B^A) = 1
Let us start with the definition of conditional probablity shown in Eq. (14.1)
Let X = A,B
Similarly, we can write
Multiply both sides with to get
From (1) and (2), we see
To do the second part, we take advanatge of the preceding result.
Rewriting the above result by flipping A and B, we get
Because the left sides of Eq (3) and (4) are the same, we can equate the right sides.
Now divide both sides with P(B|E), to get
Now replace E with C and you see the asnswer.
If B arrives first and then C, we have
If C arrives first, then we have
We need to show that the two right hand sides are equal. We can achieve this objective by applying Bayes' rule to the numerator and denominator terms in the square bracket in the second equation. Our goal is to reverse the roles of B and C. To achieve this reverseal, we need to appl y the generalized form of Bayes' Rule conditioned on A (See Prob. 14.5b, done above). Let us re-write the above equation
Cancelling P(C) and P(C|A) and re-arranging terms, we get