# Basic probability exercise

1. Feb 24, 2008

### broegger

Hi. I'm a math instructor and this problem is given to 1. year economics students.

1. The problem statement

In an episode of the TV show "All in the Family", Mike claimed that he could identify different brands of cola by taste alone. He was challenged and presented with three glasses, one filled with Coke, one filled with Pepsi, and one filled with RC Cola. Suppose that Mike really was not able to discriminate among the brands by taste.

(a) Find the probability that none of his identifications would be correct.

(b) Find the probability of no matches, had there been four brands instead of just the three.

2. Relevant equations

This should be solved using basic probability theory.

3. The attempt at a solution

By counting combinations, I have found 1/3 in (a) and 3/8 in (b), but can anyone come up with a clever, intuitive way of solving this?

Thanks.

2. Feb 24, 2008

### Defennder

How about a probability tree diagram? Since you're looking to explain basic prob theory to students it'll help if you draw out the probability tree, and then do the counting method to show that they both yield the same answer. You could probably also show them that the 2 methods are equivalent.

3. Feb 24, 2008

### broegger

Yeah, I think I'll do that, although it gets rather complex in (b).

A slight problem though: when I count the combinations I get 3/8 in (b), but when I calculate it using the multiplication rule I get 1/4

4. Feb 24, 2008

### Defennder

How did you get 3/8? I got 1/4 for both. I did it by marking out (circling or crossing) all the branches where that one guess is correct. Then I calculated the probability associated with each path and then sum them all up. It's quite tedious so it's easy to make mistakes here. My working was like 3x(1/4 x 1/3 x 1/2 + 1/4 x 1/3 x 1/2). I multiplied by three because the expression in the brackets correspond to the possible combinations to get all 4 wrong along 1 of the probability paths. And there are 4 possibilities at the start, and only 3 are wrong, so that's where 3x comes in.

5. Feb 24, 2008

### broegger

Hmm. In (a) there are two ways of getting all assignments wrong (if the correct one is 123, then these are 312 and 231) and there are 3*2*1 = 6 possible assignments. If his guesses are random this yields Pr(all wrong) = 2/6 = 1/3. In (b) I count 9 ways of getting all assignments wrong out of 4*3*2*1 = 24 possible assignments, so Pr(all wrong) = 9/24 = 3/8.

Thanks for helping out :)

6. Feb 24, 2008

### Defennder

Wow your method is even more tedious than mine since you're counting out all the exact possibilities. This will take some time. I'll post all the combinations if I can figure them out.

EDIT: This is very odd. I figured out that all the possible combinations for the 4 colas are below. Assume 1234 is the correct combination:

2143
2341
2413
3412
3421
3142
4123
4312
4321

So there are indeed 9 wrong possibilities and the total number of possibilities is 4! And 9/4! = 3/8 which is not the same as 1/4. Ugh, there must be some mistake here but I can't see it. Anyone?

Last edited: Feb 24, 2008
7. Feb 24, 2008

### broegger

:-) Precisely. Hopes somebody weighs in...

8. Apr 14, 2009

### tim thomas

The 3/8 is correct. When doing the calculation, you didn't take in to account the possibility that one of the remaining possible incorrect choices after the first might have already been used.

Using your sample space list, first find the probability of NOT picking 1 first (3/4).
Then mark off all outcomes where 1 occurs first and of those remaining find the probability of NOT picking 2 second (7/9).
Mark off all outcomes where 2 occurs second and of those remaining find the probability of NOT picking 3 third (11/14).
Mark off all outcomes where 3 occurs third and of those remaining find the probability of NOT picking 4 fourth (9/11).

First - 3/4
Second - 7/9
Third - 11/14
Fourth - 9/11

Multiply these, you get 3/8.