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A standard balanced 6 sides 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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(a) Probability roll is odd is $1/2$, even is $1/2$. Probability $5$ odd $5$ even is $\binom{10}{5}/2^{10}$, like tossing coin. – André Nicolas Sep 24 '11 at 17:27
(b.) Expectation is correct. Again, rolling a three is like tossing an unfair coin, which shows up heads with probability $1/6$. So the number of threes follows a binomial distribution $B(10, 1/6)$. Can you calculate its mean and variance? – Srivatsan Sep 24 '11 at 17:32
(c.) The events are not independent. Rolling 5 twos lowers the chance that you roll (exactly) 2 sixes. To see this, consider the related problem: Find the probability that exactly 6 sixes are rolled given that 5 twos are rolled. You can check that this probability is $0$, since rolling 5 twos automatically implies that you could've rolled at most only 5 sixes. – Srivatsan Sep 24 '11 at 17:34
(b) Mean fine. For variance use $E(X^2)-(E(X))^2$, use same technique to find $E(X^2)$. (c) Very much not independent. If lots of $2$'s, sixes less likely. And remember about the neither $2$ nor six. Basic strategy OK, probability calculations incorrect, both top and bottom. – André Nicolas Sep 24 '11 at 17:36
I always mess up on independence. I need to get a better grasp on it. – lord12 Sep 24 '11 at 17:37
up vote 1 down vote accepted

(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. – lord12 Sep 25 '11 at 0:17

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