Well done for getting the answers. Too late but here's my way. See diagram.

Bob

]]>Tnx..]]>

a)

Using a gf:

P(at least 1 six) = 1 - 25 / 36 = 11 / 36

b) Enumeration is fine but for a gf, here is one.

The ways to make an 8 are 2sx^2 and 3x^8. So the answer is 2 / 5.

]]>(b) 2/5

I am absolutely sure about the first one, but have a slight nagging doubt about the second.

The thing that occurs to me is Bayes Theorem, and I haven't studied that for a long time.

I therefore decided to think about how many possible combinations added to 8:

6,2

5,3

4,4

3,5

2,6

Then noticed that two of them contained a six. So (2/5)

The thing that made me think of Bayes was that it was "something given something else", but I cannot think how

Bayes could be used in this example. Perhaps it is ((2/36)/(5/36))

Bayes formula would be: P(K|C) = P(KC)/P(C)

So would it be valid to let P(KC) be the chance of getting an 8 which also contains a six ?

With P(C) as the chance of getting a total of 8.

With P(K) as the chance of the role containing a six.

With P(K|C) as the chance of the role containing a six given that the total is 8.

(Someone needs to check all of that because I am not sure of it. Haven't done this for ages.)

]]>2 number cubes are rolled. a)What is the probability of getting 6 at least on one cube? b)What is the probability of getting a number 6 if we know that the sum of two numbers is 8?]]>