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By Ross S., Weatherwax J.L.

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When we do so we find we obtain 672. Part (c): In the same way as in Part (b) of this problem lets count first the number of clubs with C and D in them and second the number of clubs without C and D in them. This number is 8 8 + = 64 . 1 3 Again multiplying by 3! we find a total number of 3! · 64 = 384 clubs. Part (d): For E to be an officer means that E must be selected as a club member. The 9 = 36. Again multiplying number of other members that can be selected is given by 2 this by 3! gives a total of 216 clubs.

Terms like (R, R)) and it becomes {(R, G), (R, B), (G, R), (G, B), (B, R), (B, G)} . Problem 2 (the sample space of continually rolling a die) The sample space consists of all possible die rolls to obtain a six. For example we have {(6), (1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (1, 1, 6), (1, 2, 6), · · · , (2, 1, 6), (2, 2, 6) · · · } The points in En are all sequences of rolls with n elements in them, so that ∪∞ 1 En is all c possible sequences ending with a six. Since a six must happen eventually, we have (∪∞ 1 En ) = φ.

Following the hint we let En be the event that a five occurs on the given by 1 − 91 − 16 = 13 18 n-th roll and no five or seven occurs on the n − 1-th rolls up to that point. Then P (En ) = 13 18 n−1 1 , 9 since we want the probability that a five comes first, this can happen at roll number one (n = 1), at roll number two (n = 2) or any subsequent roll. 4 . = 13 9 1 − 18 5 Problem 26 (winning at craps) From Problem 24 we have computed the individual probabilities for various sum of two random dice.

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A solution manual for A first course in probability by Ross S., Weatherwax J.L.


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