# A solution manual for A first course in probability - download pdf or read online By Ross S., Weatherwax J.L.

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For two general sets A and B the inclusion-exclusion identity is P (A ∪ B) = P (A) + P (B) − P (A ∩ B) . Using this we can evaluate the probabilities of union of events. 32 . We will now use these results in the following wherever needed. Part (a): The requested proportion of people who read only one paper can be represented from three disjoint probabilities/proportions: 1. P (I ∩ ¬II ∩ ¬III) which represents the proportion of people who only read paper I. 2. P (¬I ∩ II ∩ ¬III) which represents the proportion of people who only read paper II.

So the total number of ways to select a committee of six people is 3 3 8 4 7 2 + 8 3 7 3 Problem 13 (the number of different art sales) Let Di be the number of Dalis collected/bought by the i-th collector, Gi be the number of van Goghs collected by the i-th collector, and finally Pi the number of Picassos’ collected by the i-th collector when i = 1, 2, 3, 4, 5. Then since all paintings are sold we have the following constraints on Di , Gi , and Pi , 5 5 Di = 4 , 5 Gi = 5 , i=1 i=1 Pi = 6 . i=1 Along with the requirements that Di ≥ 0, Gi ≥ 0, and Pi ≥ 0.

Problem 17 (a simple combinatorial identity) To show that n 2 = k 2 + k(n − k) + n−k 2 for 1 ≤ k ≤ n , is true, begin by expanding the right hand side (RHS) of this expression. Using the definition of the binomial coefficients we obtain (n − k)! k! (k − 2)! (n − k − 2)! k(k − 1) (n − k)(n − k − 1) = + k(n − k) + 2 2 1 2 k − k + kn − k 2 + n2 − nk − n − kn + k 2 + k = 2 1 2 n −n . = 2 RHS = Which we can recognize as equivalent to n 2 = n 2 since from its definition we have that n! n(n − 1) = . (n − 2)!
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