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A standard balanced 6 sided die is rolled is rolled 10 times.

a. Find the probability that five of the rolls are odd numbers and five of the rolls are even numbers.

P(odd) = (10 choose 5)(1/2)^(5)(1/2)^(5) = P(even)

b. Find the expected value and variance of the number of threes rolled.

Let Ii be an indicator variable indicating whether a 3 is rolled. Then E[I1 +...i10] = 10(1/6) = 10/6.

c. Find the probability that exactly two sixes are rolled given five twos are rolled.

These are independent events so P(2 6's rolled|5 2's rolled) = P(2 6's rolled and 5 2's rolled)/P(5 2's rolled) = (10 choose 2)(1/6)^(2)(5/6)^(8).

Are these answers correct?

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    Would the answer for part c be (5 choose 2)(1/6)^2(5/6)^(3) since you only have$5$slots to put the 2's?2011-09-24

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(a) is fine. (b) has the right mean but doesn't give the variance. (c) is wrong.

I'm not sure how to interpret (c): is it that 5 dice are known to be two, or that the number of twos is exactly 5? In the latter, this is like asking about 10-5=5 dice with 6-1=5 sides each. In the former, you need to remove the 934926 cases in which at least 5 twos are rolled.

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    (5 choose 2)(1/6)^2(5/6)^(3) since you only have 5 slots to put the 2's? I think the interpretation is the latter.2011-09-25