ECS170 |
Homework Assignment |
Winter 2003 |

- Do problem 14.1 (10 points)

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

- Do problem 14.5 (20 points)

Let us start with the definition of conditional probablity shown in Eq. (14.1)

Let X = A,B

(1)

Similarly, we can write

Multiply both sides with to get

(2)

From (1) and (2), we see

(3)

QED

To do the second part, we take advanatge of the preceding result.

Rewriting the above result by flipping A and B, we get

(4)

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.

- Do problem 14.8

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