College Math

College Math Study Guide

4. c Solution: It will be really tough and monotonous to find the probability for one, then two, and then three items, and so on, such that one is chocolate. Here, we use the case of complement. That is let us find the probability that none of the item drawn is a chocolate and we deduct it from the total probability. The total elements for drawing no chocolates (that all items drawn are candies) would be, C (15, 5) = 3,003 The total sample space would be found by C(25, 5) = 53,130 Hence, the probability for drawing no chocolates would be: P(E) = 3,003/53,130 = 0.0565 Now, the probability of drawing at least one chocolate would be: P(E’) = 1 – 0.0565 = 0.9435 = 94% 5. a Solution: We use the formula of expected value in this case. The two events are arrow stopping at yellow and orange colors, respectively. The payoff for yellow (Event a 1 ) is -2 and the payoff for orange (event a 2 ) is 3 (that is 5-2). We must calculate the probabilities for both of these events. The sample space is 10 (since there are 10 regions), and number of elements in winning regions n(a 2 ) = 4 and number of elements in losing regions n(a 1 ) = 6. The probability of event a 1 = 6/10 = 0.6 = p 1 The probability of event a 2 = 4/10 = 0.4 =p 2 The expected payoff would be: = + E = -2* 0.6 + 3*0.4 = 0

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