1.) Given two urns one with three red and three white balls and the other with three pink, one yellow and four red balls.  If one draws a ball fromeach urn, estimate the probability of the following outcomes:
 

Two red balls

Using the unrestricted conjunction principle (events are independent) you multiply the probability of drawing a red from urn one (.5) by the probability of getting a red drawing from urn two (.5).

.5 x .5 = .25
 

One white ball and one yellow ball

Using the unrestricted conjunction principle (events are independent) you multiply the probability of drawing a white ball from urn one (1/2) by the probability of drawing a yellow ball from urn two (1/8).

1/2 x 1/8 = 1/16 or .0625
 

One white or one pink

Using the restricted disjunction rule (the events are not mutually exclusive) you add the probability of getting a white ball from urn one (1/2) and the probability of getting a pink ball from urn two (3/8) and subtract the probability of getting both (1/2 x 3/8).

(1/2 + 3/8) - (1/2 x 3/8) =
(4/8 + 3/8) - 3/16 =
7/8 -3/16 =
14/16 - 3/16 = 11/16
 

2.) Bill believes that the odds of his brother not bribing a theater owner are .96.  However, Bill discovers that Bob got tickets for the opening of the Harry Potter.  The odds of getting tickets if you don't bribe the theater owner are .1.  The odds of getting tickets if you bribe the theater owner are .6.  What should Bill believe about the likelihood of his brother bribing a theater owner now?

Using Bayes theorem .96 is the odds of Bob not bribing a theater owner.  The odds of his getting a ticket without bribing is .1.  The odds of Bob bribing a theater owner are .04.  The odds of Bob getting tickets if he did are .6.  Thus, the new odds that Bob bribes are .8 given that he got tickets.

3.) What are the odds of getting a 6 and a 1 or a 6 and a 3 or a 6 and a 6 on two rolls of a dice.

There are two ways of getting each of the first two combinations.  For example, you can get a 6 and a 1 by rolling a 6 first and a one second, or a 1 first and a 6 second.  The odds of getting any number on a roll is 1/6.  Thus, using the restricted conjunction rule the odds of getting a 6 then a 1 is (1/6 x 1/6) = 1/36.  The odds of getting a 1 and then a 6 is also (1/6 x 1/6) = 1/36.  The same for rolling a 3 and a 6.   There is only one way to get a 6 and a 6.  Thus, to solve the problem you need to calculate the odds of getting each combination using the restricted disjunction rule.

6 and 1 = 1/36 + 1/36 = 2/36 = 1/18
6 and 3 = 1/36 + 1/36 = 2/36 = 1/18
6 and 6 = 1/36 = 1/36 

Then you must calculate the odds of getting one or the other of the combinations using the restricted disjunction rule:

2/36 + 2/36 + 1/36 = 5/36