- Suppose A is the event of getting at exactly one ace when dealt 5 cards from an ordinary deck of cards. Suppose
B is the event of getting all hearts when dealt 5 cards. Compute the following probabilities. For questions a-d do
not simplify the arithmetic computations. For e you need to do the calculation.

a) P(A)

b) P(B)

c) P(A|B)

d) P(A&B) (you may use Bayes Rule)

e) Are A and B independent?

- From the text (new edition) pg 489, do problem 13.6. Be careful about the difference between BOLD P (a vector) and non-bold P. Here is the problem. Using the joint probability shown in Figure 13.3 (handed out in class) compute:

a) P(toothache)

b)**P**(Cavity)

c)**P**(Toothache | cavity)

d)**P**(Cavity | toothache or cavity)

- From the text (new edition) pg489, do problem 13.8. For convenience, I've rewritten the problem. Be careful that you use all the information.

After your yearly checkup, the doctor has bad news and good news. The bad news is that you tested positive for a serious disease and that the test is 99% accurate (i.e., the probability of testing positive when you do have the disease is 0.99, as is the probability of testing negative when you don't have the disease). The good news is that this is a rare disease, striking only 1 in 10,000 people. Why is it good news that the disease is rare? What are the chances that you actually have the disease?

Hint: Besides Bayes Rule, recall that P(A)+P(not A) = 1. This extends to P(X|Y)+P(not X|Y) =1.