What is the probability that the sum is 11, given that at least one of the numbers rolled is even?

Question

involving the rolling of two fair dice. Answer the following questions.
(1) What is the probability that the sum is 11, given that at least one of the numbers rolled is even?

(2) What is the probability that at least one number is even, given that the sum is 11?

(3) What is the probability that at least one number is even, given that the sum is 10?

Answer 1

1.  I did this by tabulating all possible combinations, and then deleting the one with no even numbers.  This leaves 27 combinations, 2 of which add up to 11.  Thus p = 2/27.  I'm not certain how to do it another way....

 

2.  there are only 2 combinations that give a sum of 11, and both have even numbers.  p = 1

 

3.  there are only 3 combinations with a sum of 10, and 2 of them have even numbers.  p = 2/3

 

Because of the way I did this, you should check using whatever rules you have learned.

Answer 2

Katie,

This is a set of all possible outcomes

1 - 1 - 2 - 3 - 4 - 5 - 6, 3 outcomes where one is even

2 - 1 - 2 - 3 - 4 - 5 - 6, 6 outcomes where at least one is even

3 - 1 - 2 - 3 - 4 - 5 - 6, 3 outcomes where one is even

4 - 1 - 2 - 3 - 4 - 5 - 6, 6 outcomes where at least one is even

5 - 1 - 2 - 3 - 4 - 5 - 6, 3 outcomes where one is even

6 - 1 - 2 - 3 - 4 - 5 - 6, 6 outcomes where at least one is even

                                  27 outcomes where at least one is even

1 - 1 - 2 - 3 - 4 - 5 - 6, 

2 - 1 - 2 - 3 - 4 - 5 - 6, 
3 - 1 - 2 - 3 - 4 - 5 - 6, 
4 - 1 - 2 - 3 - 4 - 5 - 6, 
5 - 1 - 2 - 3 - 4 - 5 - 6, 
6 - 1 - 2 - 3 - 4 - 5 - 6,  the only two outcomes where the sum is 11 (5,6),(6,5)=> p(1)=2/27

p(2)=1 (100%), the only way you get 11, adding 5 & 6, involves one even (6) and one odd (5)

p(3)=2/3 all possible combinations are {(4,6),(6,4),(5,5)}

I hope this helps

Answer 3

The basic thing you are trying to find is (how many ways can you get the desired outcome) over (total possible outcomes)  or  D/O  where D is the outcome you want and O is the total possible outcomes.

 

(1) If one die is an even number then that is only 3 possibilities on one die... it can be 2, 4, or 6

The other die can be any number from 1 through 6

 

If one die is a 2 then the possible sums are 3,4,5,6,7,8  (6 possible outcomes)

If one die is a 4 then the possible sums are 5,6,7,8,9,10 (6 possible outcomes)

If one die is a 6 then the possible sums are 7,8,9,10,11,12 (6 possible outcomes)

 

there are 18 possible outcomes in the scenario as stated so O = 18

Only one of those outcomes is the desired result of 11 so    D = 1

 

Thus the probability of rolling a sum of 11 given that one of the numbers rolled is even is 1/18

 

(2) The only possible way to get a sum of 11 is to roll either a 5,6 or a 6,5

     Given that the sum must be 11 then the probability of one number being even is 100% or P=1

 

(3) If one does not eliminate doubles... so that rolling a 1 then a 6  or (1,6) is considered different

      from rolling a 6 then a 1 or (6,1) even though both total 7

 

      the only way to get a sum of 10 is to roll either (4,6), (5,5), or (6,4).... 3 ways to get a sum of 10

      2 out of those 3 have at least one even number (4,6) and (6,4)

      Thus 2 out of 3 is the probability that at least one number is even given that the sum is 10

       P = 2/3