Probability questions...?

A) There are six faces on a die. The probability that the first one is **not** a three is 5/6. The probability that the second one is not a three is also 5/6. So the probability that neither of them is a 3 is (5/6) * (5/6) = (25/36)

Getting at least one three is the opposite of getting no 3s so the probability of getting at least one 23 is (1) - (25/36) = (36/36) - (25/36) which is equal to:

11/36.....<<<...Answer to (A)

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B) There are 36 permutations of rolling two dice. 6^2 = 36, so there are 36 ways to roll the two dice (just a pseudo-proof).

A seven can be rolled as n followed by 7-n for each side n. That is 6 ways.

Therefore the probability of rolling a 7 is 6/36 =

1/6.....<<<...Answer to (B)

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C) The probability of one die coming up odd is 2/6 = 1/3. The probabiltiy that they are both odd is:

(1/3) * (1/3) = (1/9).....<<<..Answer to (C)

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a) Independent

That is because the probability of rolling a seven was originally 1/6, and after the roll of a 3 it is still 1/6

They are not mutually exclusive - you can first roll a 3 and then a 4 totaling 7.

b) These are dependent results. Originally the probability of rolling two odd dice was 1/9, but if you first roll a three the probability that the next one is odd. Since 1/3 not= 1/9 they are dependent.

They are clearly not mutually exclusive. First a 3, then a 5 satisfies (A) and (C)

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