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

By Ross S., Weatherwax J.L.

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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))).

5 Part (c): To have three people between A and B, A and B must be on the ends with 3! = 6 possible ordering of the remaining people. Thus with two orderings of A and B we have a probability of 2·6 1 = . 5! 10 Problem 45 (trying keys at random) Part (a): If unsuccessful keys are removed as we try them, then the probability that the k-th attempt opens the door can be computed by recognizing that all attempts up to (but not including) the k-th have resulted in failures. Specifically, if we let N be the random variable denoting the attempt that opens the door we see that P {N = 1} = P {N = 2} = P {N = 3} = 1 n 1 n 1 1− n 1− 1 n−1 1− 1 n−1 1 n−2 1− 1 n−1 ··· 1− ..

Since there are two possible classes to select this student from this can be done in two ways. Once both of these classes are selected we pick the individual two and one students from their n n ways respectively. Thus in total we have and respective classes in 1 2 3·2· n 2 n 1 = 6n n(n − 1) = 3n2 (n − 1) , 2 ways. Part (d): Three students (all from a different class) can be picked in Part (e): As an identity we have then that 3n 3 =3 n 3 + 3n2 (n − 1) + n3 . n 1 3 = n3 ways. We can check that this expression is correct by expanding each side.

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