# Thread: dice , conditional probability

1. ## dice , conditional probability

could any 1 help me answer this question?

When two dice were rolled, it is known that the sum was an even number. In this case, find the probability that the sum was 8.

tx alot

2. $\displaystyle P(\text{sum was 8}|\text{sum was even}) = \frac{P(\text{sum was 8} \cap \text{ sum was even})}{P(\text{sum was even})}$ $\displaystyle = \frac{P(\text{sum was even}|\text{sum was 8})P(\text{sum was 8})}{P(\text{sum was even})}$

3. so what would the answer be in a fraction ?

two dice ,rolling 2 sixes

1/36

etc

4. In tossing two dice, we get 36 possible pairs.
How many of those pairs give an even sum? (hint: both even or both odd).
How many of those pairs sum to 8?

5. 5-1 , 1-5 , 6-2 , 2-6 , 4- 4

so 5/18 i think this right

this may seem like a silly quesion to some

4/9

6. Here's the very basic way to so it:

The totals on two dice and the different ways they can occur are:
2: 1, 1 one way
3: 2, 1; 1, 2 two ways
4: 3, 1; 2, 2, 1, 3 three ways
5: 4, 1; 3, 2, 2, 3, 1, 4 four ways
6: 5, 1; 4, 2, 3, 3, 2, 3, 1, 5 five ways
7: 6, 1; 5, 2; 4, 3, 3, 4, 2, 5, 1, 6 6 ways
8: 6, 2; 5, 3; 4, 4, 3, 5, 2, 6 five ways
9: 6, 3; 5, 4, 4, 5, 3, 6 four ways
10: 6, 4; 5, 5; 4, 6 three ways
11: 6, 5; 5, 6 two ways
12: 6, 6 one way
A total of $\displaystyle 6\times 6= 36$ ways.

The even numbers are 2: one way, 4: three ways, 6: five ways, 8: five ways, 10: three ways, 12: one way, a total of 1+ 3+ 5+ 5+ 3+ 1= 18 ways. Of those five give a total of 8: the probability that a pair of dice, given that the sum is even, have a sum of 8 is 5/18.

Once again, I didn't read all of the posts carefully- Plato has already suggested this and the OP did it properly.