First what u need to do is make a sample table of the 2 die's

like this

(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)

(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)

(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)

(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)

(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)

(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)

Let the sample space be denoted as S = 36

Now

**a)Probability of sum being atleast 9** ie Pr(X

9)

which is

Pr(X

9) = Pr(X = 9) + Pr(X = 10) + Pr(X = 11) + Pr(X = 12)

Mr F says: This is . No harm done though, the OP just needs to subtract from 1 the result of the calculation below.
=

= .....rest i will leave up to u to compute

**b)Probability of sum less than 4 or greater than 9 **
Take Pr(A) = Probability of sum being less than 4 ie(1,1) (1,2) (2,1)

Pr(A) =

=

Take Pr(B) = Probability of sum being greater than 9 ie. (4,6) (5,5) (6,4) (5,6) (6,5) (6,6)

Pr(B) =

Now question is prob of sum less than 4 or greater than 9 and here or means

Pr(A

B) = Pr(A) + Pr(B) .......since they both are mutually exclusive

=

= .............and u can do it from here on by urself

**c)Probability that the sum is atleast 10, given that the sum is atleast 9**
here Pr(A) = sum atleast 10...........which we need to find out

Pr(B) = given that the sum is atleast 9

Pr(A

B) = reading which are both in Pr(X

9) and when Pr(X

10) ie (4,5) (5,5) (5,4) (6,5) (5,6) (6,6) = .........

=

there using conditional prob

Pr(A|B) =

= ...........substitute and get the answer

**d) the sum is more than 7, given that the sum is no more than 5.**
It is also quite similar to the c) .........so u should be able to do it