- #1
CAF123
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Homework Statement
In successive rolls of a pair of fair die, what is the probability of getting 2 sevens before 6 even numbers?
The Attempt at a Solution
So, I read this once and then twice and then a third time. You can't get a seven on a fair die (assuming the die is six sided)? I proceeded on the assumption the question means the sum of the two die is seven and sum of two die is even
Let E be the event of rolling seven: E = {(1,6),(2,5),(3,4)} => P(E) = 3/21 = 1/7.
I believe the size of the sample space, |s| = 21, since we don't want to distinguish between, say, (1,6) and (6,1).
Let F be the event that you get 6 even numbers: F = {(1,1),(1,3),(1,5),(2,2),(2,4),(2,6),(3,3),(3,5),(4,4),(4,6),(5,5),(6,6)} => P(F) = 12/21.
Given that the number of rolls can be infinite, P(2 sevens) = 1 and P(6 even numbers) = 1, since there is no limit on the number of rolls and if something can happen, it will happen. Using the eqn P(2 sevens before 6 evens) = P(2 sevens)/(P(2 sevens) + P(6 evens)) gives an answer of 1/2, which is close to the actual answer. Why does this not work?
Also, I am aware the above is quite a 'mechanical' way of doing this question. I tried it using summations with the probabilities I got above (1/7 and 12/21), but I don't get far.
Many thanks for any advice.