By Ross S., Weatherwax J.L.

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**Sample text**

64 − 8)! 64 8 = 4426165368 . The number of locations where eight rooks can be placed who won’t be able to capture any of the other is given by 82 · 72 · 62 · 52 · 42 · 32 · 22 · 12 , Which can be reasoned as follows. The first rook can be placed in 64 different places. Once this rook is located we cannot place the next rook in the same row or column that the first rook holds. This leaves seven choices for a row and seven choices for a column giving a total of 72 = 49 possible choices. Since the order of these choices does not matter we will need to divide this product by 8!

Problem 12 (language probabilities) Let S be the event that a student is in a Spanish class, let F be the event that a student is in a French class and let G be the event that a student is in a German class. 02 . 06 Part (a): We desire to compute P (¬(S ∪ F ∪ G)) = 1 − P (S ∪ F ∪ G) . 5 . 5. Part (b): Using the definitions of the events above for this subproblem we want to compute P (S ∩ (¬F ) ∩ (¬G)) , P ((¬S) ∩ F ∩ (¬G)) , P ((¬S) ∩ (¬F ) ∩ G) . As these are all of the same form, lets first consider P (S ∩ (¬F ) ∩ (¬G)), which equals P (S ∩(¬(F ∪G))).

Problem 39 (five different hotels) When the first person checks into the hotel, the next person will check into a different hotel with probability 54 . The next person will check into a different hotel with probability 53 . 48 . 5177 . Problem 42 (double sixes) The probability that a double six appear at least once is the complement of the probability that a double six never appears. The probability of not seeing a double six is given by 1 = 35 , so the probability that a double six appears at least once in n throws is given 1 − 36 36 by n 35 1− .

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

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