As it is in the book

See book

Note: Think of this in terms of a game. I give you the two values $p_/alpha$ and $p_/beta$ and you give me the max and min for the intersection and union. Test your answer, by trying some values.

Note that X, Y and Z are *SETS* of variables. So interpret $P(\chi)$ as the joint probability of the union of all of these sets of variables. So if X={A,B} Y={C} and Z={D,E} where A,B,C,D,E are all rv's $P(\chi)$ is P(A,B,C,D,E), the joint over all of the rv's, not some sort of probability over set-valued random variables.

I think this one is clear as it is, let me know if you think not.

A deck of 52 cards contains 4 aces. If the cards are shuffled and distributed in a random manner to 4 players so that each player receives 13 cards, what is the probability that each player receives 1 ace?

This problem comes from G. Gigerenzer, “Calculated Risks: How To Know When Numbers Deceive You”, Simon and Schuster Press, 2002. Give the answer and show how you obtain the results using Bayes rule. This is the classic use of Bayes' law.

To diagnose colorectal cancer, the hemoccult test — among others — is conducted to detect occult blood in the stool. This test is used from a particular age on, but also in routine screening for early detection of colorectal cancer. Imagine you conduct a screening using a hemoccult test in a certain region. For symptom-free people over 50 years old who participate in screening using the hemoccult test, the following information is available for this region.

The probability that one of these people has colorectal cancer is 0.3 percent. If a person has colorectal cancer, the probability is 50 percent that this person will have a positive hemoccult test. If a person does not have colorectal cancer, the probability is 3 percent that this person will still have a positive hemoccult test. Imagine a person (over age 50, no symptoms) who has a positive hemoccult test in your screening. What is the probability that this person actually has colorectal cancer.

Let $P(x) = 0.2$ for a Boolean Random Variable X.

Suppose that X influences another Random Variable, Y, in the following way:

$P(Y=1|x) = 0.2\,$

$P(Y=2|x) = 0.4\,$

$P(Y=1|\neg x) = 0.6\,$

$P(Y=2|\neg x) = 0.3\,$

Note that Y can take three values 1,2, or 3

Compute the conditional probabilities:

Note that we will be computing many probabilities in this and subsequent sections. You may compute the joint of x and y and then sum the needed Values out of your joint or use Bayes' law directly.

1. $P(x|Y=1)\,$
2. $P(x|Y=2)\,$
3. $P(x|Y=3)\,$