1. Apr 15, 2012

### kenny1999

1. The problem statement, all variables and given/known data

The question below, i already have the answer. but I can't work it out. Can anyone help?

Questions (Original)

In a game, Tom throws a fair dice twice. 1 point is awarded for each '6' thrown, 2 points are awarded if the sum of the two numbers thrown is prime, and no points are awarded for other outcomes. Find the probabilities that

(a) Tom gets 1 point in the game.
(b) Tom gets 2 points in the game.
(c) Tom gets at least 2 points in the game.

2. Relevant equations

3. The attempt at a solution

(a) There are total 36 possible results for a fair dice to be thrown twice. For each '6' thrown 1 point is awarded so there are (1,6) (2,6) (3,6) (4,6) (5,6) (6,6) and (6,1) (6,2) (6,3) (6,4) (6,5), but (6,6) is not counted because it gives two points, and (1,6) and (5,6) are also left out because they give prime number on sum.

So the desired outcomes should be (2,6)(3,6)(4,6)(6,4)(6,3)(6,2), then 6/36 = 1/6 but the answer is wrong. I am not going to say the answer but I just don't understand.

Please also explain (b) and (c) if possible. Thank you.

2. Apr 15, 2012

### Ray Vickson

My answer for (a) is different from yours, but that is because I was careful to count things correctly. Of course, for (b) and (c) I would just find and list all the relevant outcomes that have the desired point scores. That is what you should do as well.

RGV

3. Apr 15, 2012

### kenny1999

why not 1/6?

(2,6) (3,6) (4,6) (6,4) (6,3) (6,2) are all counted as 1 point because sum of their numbers doesn't give a prime number. (1,6) (5,6) (6,5) as well as (6,1) are not counted because they get 3 points.

I don't think there are any other outcomes that would give 1 point only in addition to my count. So why not 1/6? Please tell me.

4. Apr 15, 2012

### Ray Vickson

Sorry, I mis-read the question. Your analysis of (a) is correct, so I suppose the answer in the book is wrong. That happens a lot, by the way.

RGV