a pair of dice are rolled: find the following..?

Question

a) probability of a sum greater than 3

b) probability of a sum NOT equal to 9 or 10

c) probability of a sum that is even and less then five


thanks!!

Answer 1

You have six choices for the first die and six for the second die. That's a total of 36 possible outcomes for the two dice. The sum of the dice will be between 2 and 12.

PROBLEM A:

Rather than figuring out all the possible ways to get sums greater than 3, figure out the ways to get sums less than or equal to 3 and then subtract from 1.

The probability of a sum greater than 3 is equal to:
1 - P(sum <= 3)

P( sum = 2) --> (1+1) --> 1 ways out of 36
P( sum = 3) --> (1+2) or (2+1) --> 2 ways out of 36

So the P ( sum <= 3 ) = 1/36 + 2/36 = 3/36 = 1/12

P( sum > 3 ) = 1 - P( sum <= 3 )
= 1 - 1/12
= 11/12

PROBLEM B:

Similar method. Figure the ways to get 9 or 10, then subtract from 1.

P( sum = 9 ) --> 3+6, 4+5, 5+4, 6+3 --> 4 ways out of 36
P( sum = 10 ) --> 4+6, 5+5, 6+4 --> 3 ways out of 36
P( sum = 9 or 10 ) = 4/36 + 3/36 = 7/36

P( sum not 9 or 10 ) = 1 - 7/36 = 29/36

PROBLEM C:

The even sums less than 5 are 2 and 4:

P( sum = 2 ) --> 1+1 --> 1 way out of 36.
P( sum = 4 ) --> 1+3, 2+2, 3+1 --> 3 ways out of 36

P( sum is even and less than 5 ) = 1/36 + 3/36 = 4/36
= 1/9

Answer 2

First note the total number of possibilities is 36 = (6*6)

a)
What is the probability of rolling a sum not greater than 3?
There are 3 such ways (1,1) (2,1) or (1,2)
So thats 3/36
We want the complement so we take 1-3/36 = 33/36 = 11/12 the required answer.
b)
Again what is the probability that it equals 9 or 10?
(3,6) (4,5) (5,4) (6,3) (4,6) (5,5) (6,4) So that 7 out of 36 ways
take 1-7/36 = 29/36 the desired answer.
c)
this one we can count directly: The sum must be 2 or 4 so:
(1,1) (1,3) (2,2) (3,1) Thats 4/36 or 1/9 as required

Hope this helps!