EECS 391/491: PROBLEM SET #7 FAQ

Overall

Conditional Probability Example

Consider the following table for the total of throwing two dice:

Two dice totals

Die 1

Die 2

1

2

3

4

5

6

1

2

3

4

5

6

7

2

3

4

5

6

7

8

3

4

5

6

7

8

9

4

5

6

7

8

9

10

5

6

7

8

9

10

11

6

7

8

9

10

11

12
[Taken from http://wizardofodds.com/gambling/dice.html]

As an example, here are some probabilities and conditional probabilities computed with the help of this table: 

P(Die 1 is odd) = 18/36 = 1/2
P(Die 2 is odd) = 18/36 = 1/2

EVENT A: at least one of the two are odd
EVENT B: total is odd
EVENT C: total is even

P(A) = (18+9)/36 = 27/36 = 3/4
P(A|B) = P(A, B)/P(B) = (18/36)/(18/36) = 1
P(A|C) = P(A, C)/P(C) = (9/36)/(18/36) = 1/4

"Rule of Sum and Product"

The rule of sum and product are simple principles that are useful for counting (and, hence, for computing probabilities): 
Given mutually exclusive sets X_i, each
with a number of elements |X_i|=m_i, then

(Rule of Sum) the number of ways to make a choice from
X_1 or X_2 or ... or X_n is m_1 + m_2 + ... + m_n;

(Rule of Product) the number of ways make a choice from each of
X_1 and X_2 and ... and X_n is m_1 * m_2 * ... * m_n

For example, I have 16 books on Java and 4 books on Perl (and none that cover both). The "rule of sum" says there are 20 books I could choose from to learn about either of the two subjects. If I must learn about both Java and Perl, and choose one of each, then the total number of two-book libaries is 16*4.

Problem 7.5

For the first part, the joint distribution need not be given explicitly as a table. You can state the probabities w/o explicitly representing them. For instance, chance of any card in a deck is 1/52. Chance of winning big jackpot in a pick 6 of 40 lottery is 1 /[ 40 choose 6]. We did more examples in class for poker hands.

For the second part, recompute the general probabilities (recount) to be consistent with all the data given in the figure.

Author: Michael S. Branicky