You roll two standard number cubes. What is the probability that the sum is even, given that one number cube shows a 2?
did you list all the possibilities pairs that has 2?
no....
well are you going to?
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 so probability of at least one cube shows a 2 is 11/36.
It says "given that one number cube shows a 2", so let's assume that it's the first number cube that shows a 2. If that's the case, then we look at row 2 which is 2,1 2,2 2,3 2,4 2,5 2,6 now what are the sums of each pair?
3,4,5,6,7,8
which of those sums are even?
all possible pairs that has 2 is: (1,2) (2,1) (2,2) (2,3) (3,2) (2,4) (4,2) (2,5) (5,2) (2,6) (6,2) So there are 11. Out of those 11 pairs, which one has the sum of even?
(2,2) (2,4) (4,2) (2,6) (6,2)
so that would be 5/11
my book says 1/2.......
but i dont see how
If the first die is 2, then we have this set of possibilities 2,1 2,2 2,3 2,4 2,5 2,6 They add to: 3, 4, 5, 6, 7, 8 Of that list, only 4, 6, 8 are even. This is 3 sums out of 6 so, 3/6 = 1/2 is your answer
Then the question assume only ONE of the cube shows 2. Show the pair (2,2) wasn't not supposed to be counted. So answer is indeed 5/10 = 1/2
But really, the book wasn't being specific because such phrase "given that one number cube shows a 2" actually means AT LEAST one.
so i only needed 1 die with a 2 on it not both of them
Yes, in this case. Sometimes the misuse of language can change the meaning of a question. But as long as it's not a conceptual misunderstanding, then you'll be fine.
ty @jim_thompson5910 @@sourwing
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