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STP 421 - Problem Set 1 Solutions 1.) Suppose that three balls are sampled at random and without replacement from an urn containing 4 white balls and 2 black balls. Here, sampling at random means that each ball is equally likely to be chosen and that the choices of different balls are independent of one another. Sampling without replacement means that once a ball is sampled, it is permanently removed from the urn and cannot be sampled again. (a) Write down a sample space for this experiment assuming that we are only told the number of white balls in the sample. (b) Write down a sample space assuming that we also know the order in which white and/or black balls are sampled. For example, one possible outcome would be BWB, assuming that we first sampled a black ball, then a white ball and then again a black ball. Solutions. (a) If we are only told the number of white balls in the sample, then there are three possible outcomes, depending on whether the sample contains 1, 2 or 3 white balls. (Since there are only two black balls in the urn, at least one white ball must be sampled.) Accordingly, the sample space can be taken to be Ω = {1, 2, 3}. (b) If we are told the order in which white and black balls are sampled, then there are seven possible outcomes, giving Ω = {BBW, BW B, W BB, BW W, W BW, W W B, W W W }. 2.) Suppose that a pair of fair dice is rolled and that all 36 outcomes are equally likely. (a) Calculate the probability that the dice land on a pair of numbers with a sum greater than 7. (b) Calculate the probability that the both die land on the same number. Solutions. (a) If X denotes the sum of the two numbers, then P(X > 7) = P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12) 5 4 3 2 1 = + + + + 36 36 36 36 36 5 = . 12 (b) If X1 and X2 denote the two numbers rolled, then P(X1 = X2 ) = 6 X k=1 6 X 1 1 P(X1 = X2 = k) = = . 36 6 k=1 3.) Suppose that A and B are mutually exclusive events for which P(A) = 0.6 and P(B) = 0.3. What is the probability that 1 (a) either A or B occurs? (b) both A and B occur? (c) neither A nor B occurs? (d) A occurs and B does not occur? Solutions. (a) Since A and B are mutually exclusive, we have P(A ∪ B) = P(A) + P(B) = 0.9. (b) Since A and B are mutually exclusive, we also know that A ∩ B = ∅ and so P(A ∩ B) = 0. (c) Since (A ∪ B)c is the event that neither A nor B occurs, P((A ∪ B)c ) = 1 − P(A ∪ B) = 0.1. (d) Since A and B are mutually exclusive, it follows that A ⊂ B c and so P(A∩B c ) = P(A) = 0.6. 4.) A forest contains 20 elk, of which 10 are captured, tagged and then released. Some time later, 4 of the 20 elk are recaptured. What is the probability that exactly two of these are tagged? What assumptions are you making? Solution: Let us assume that no elk leave or enter the population for the duration of the study and that each set of 4 elk is equally likely to be recaptured. Since there are 20 4 = 4845 10 10 such sets and there are 2 2 = 2025 sets which contain exactly two tagged elk and two untagged elk, it follows that the probability that exactly two of the tagged elk are recaptured is 2025/4845 = 135/323 ≈ 0.418. 5.) Suppose that E and F are events and let E ∩ F denote the event that both E and F occur. Show that P(E ∩ F ) ≥ P(E) + P(F ) − 1. Solution: Since P(E ∪ F ) ≤ 1, it follows from the sum rule that P(E ∩ F ) = P(E) + P(F ) − P(E ∪ F ) ≥ P(E) + P(F ) − 1. This result is sometimes known as Bonferroni’s inequality. 2

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