g q = N log(35/36) < 0.5, then - N log(36/35) < 0.5, then N > 0.5/log(36/35). The equation can also be solved using a computer spread- sheet. Either method yields the solution N = 25, for which 1 - qN =0.5055. Drawing cards Five cards are successively drawn at random from a 32-card deck. What are the probabilities for the resulting hand to: (a) Have only red cards? (b) Include at least one ace? 1.3 Combined, joint, and conditional probabilities Answer: Recall first that a 32-card deck has four series (red hearts, red diamonds, black clubs, and black spades) of eight cards each, including an ace, a king, a queen, a jack and four cards having the values 10, 9, 8, and 7. Considering question (a), we must calculate the probability of drawing five red cards successively. We first observe that half of the cards (16) have the same (red or black) color. The probability of drawing a red card is thus 1/2, or, more formally, p(x1 = red) = C16/C32 16/32 = 1/2. For the second draw, we have 31 cards left, which by definition include only 15 red cards. Thus, p(x2 = red) = C15/C3 = 15/31, and SO on for the other three selections. The final answer for the joint probability is, therefore: p(red, red, red, red, =0.021. Question (b) concerns a hand having at least one ace. We could go through all the possibilities of successively drawing five cards with one, two, three, or four aces, but we can proceed much more quickly. Indeed, we can instead calculate the probability of the complementary event "no ace in hand," then use the property p(at least one ace) = 1 - p(no ace). The task is to use combinatorics to evaluate the probabilities of having no ace at each draw. The number of ways of drawing a first card that is not an ace is C32-4 C18 = 28. The probability of having no ace in the first draw is, therefore, 28/32. With similar reasoning for the next four draws, it is straightforward to obtain finally: pinnaco) and p(at least one ace) 0.488 = 0.511. We see that the probability of eventually having at least one ace in the five-card ha