1. Oct 9, 2008

### I_LuV_FiZiX

1. The problem statement, all variables and given/known data
Wanda was a waitress in Waves, a dining room of the Titanic. During the voyage she determined the probability that a customer ordered red wine was 0.65, and the probability that they ordered beef was 0.50. Given that they ordered red wine, the probability they ordered beef was 0.75.

a.) What is the probability that a customer orders beef and red wine?

b.) What is the probability that a customer orders beef or red wine?

c.) What is the probability that a customer orders red wine if they order beef?

d.) Are the events "orders beef" and "orders red wine" disjoint?
Give answer as yes, no, or can't tell.

e.) What is the probability that a customer orders neither red wine nor beef?

2. Relevant equations

3. The attempt at a solution
Well, the thing that's getting me is the the last sentence in the paragraph of the original information given. Is that meant to throw me off in any way? Which of these questions does it apply to?

For part a) I multiplied 0.65 by 0.5 to get my answer of 0.33.

For part b) I added the probabilty of getting red wine to the probability of getting beef and then subtracted the probability of getting both (my answer in a)) to get 0.82.

For part c) I am completely unsure as of where to even start.

For part d) I answered no, since they have no common elements influencing one another.

For part e) I did 1-(my answer in b) which is 0.82) to get 0.18

2. Oct 9, 2008

### e(ho0n3

Everything apart from c) seems correct. Concerning c), do you know how to compute conditional probabilities?

3. Oct 9, 2008

### Dick

You can only multiply probabilities to compute things like that if you know the two choices are independent of each other. In this case there is NO reason to believe that. Suppose the wine choices are Red and White, and the food choices are Beef and Fish. That gives you four cases RB, RF, WB, WF. You know the sum of them is 1.0. The other facts you are given let you write down three other equations, for a total of four equations in four unknowns. You can solve for all of the variables. What are the equations?

4. Oct 9, 2008

### I_LuV_FiZiX

So Dick, are you saying that my answers are incorrect and I have to compute them another way? I.E. they are disjoint?

5. Oct 9, 2008

### Dick

Yes, they are. Reread my post for how to do it.