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Solutions – Take-Home Exam 3 REGULAR PROBLEMS 1. Applying the counting principle here yields a sample space consisting of 2 4 16 outcomes (e.g. HHTT, TTTT, or HTHT) when flipping a fair coin 4 times in a row. The event is getting at least 2 heads, so its complementary event is getting no heads or only one head. There are exactly 5 outcomes in this complementary event: TTTT, HTTT, THTT, TTHT, and TTTH. Therefore, the probability of getting at least 2 heads is given by 5 11 1 , or 68.75%. Note: You could also use a tree diagram for this problem. 16 16 2. Note that since trifecta wagers require selecting in exact order the first 3 horses that finish the race, their probabilities involve permutations. On the other hand, since superfecta box wagers do not require selecting the first 4 horses that finish the race in exact order, their probabilities involve combinations. We thus have the following probabilities for this race: P(win TRIFECTA) 1 1 1 11 3! 8! 11! 11! 11 10 9 990 11 P3 P(win SUPERFECTA) 1 1 11 4!4! 7!4! 4 3 2 1 11! 11! 11 10 9 8 330 11C4 1 1 3 , you are exactly three times more likely to win a 990 330 superfecta box than a trifecta wager for this race. Since 3. For this slot machine we have the following probabilities: P(orange on wheel 1) P(bells on all 5 22 3 4 4 6 48 wheels) 22 22 22 10,648 1,331 20 20 21 8,400 1,050 P(no bars) 22 22 22 10,648 1,331 1 1 1 1 P(all 7' s) 22 22 22 10,648 4. 5. a) By the counting principle, there are 4! 4 3 2 1 24 ways of placing the four cards in the four boxes. b) There are 9 ways of placing these cards so that no card has the same letter as its outcomes: BADC, BCDA, BDAC, etc.) So the odds box (check this by listing these in favor of this special type of placement of the cards are 9 to 15, or 3 to 5. a) There are 11 prime number slots in Roulette: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, and 31. Thus, the odds against the ball landing on a prime number slot are 27 to 11. b) If we exclude 0 and 00 as zeros, then there are 6 slots with numbers divisible by 2 and 3 in Roulette: 6, 12, 18, 24, 30, and 36. (Note that these are the numbers that are divisible by 6 = 2 x 3). Therefore, the probability of landing on one of 3 6 . these slots is given by 38 19 c) The odds in favor of not having the ball land on a slot numbered between 1 and 36 is equivalent to the odds of having the ball land on either 0 or 00. Those odds are 2 to36, or 1 to 18. 6. Applying the counting principle here yields a sample space consisting of 6 3 216 outcomes (e.g. (1, 6, 5), (4, 4, 4), (5, 4, 3), etc.) when rolling three regular dice. A sum of 5 or less occurs only for the following 10 outcomes: (1, 1, 1), (1, 1, 2), (1, 2, 1), (2, 1, 1), (2, 2, 1), (2, 1, 2), (1, 2, 2), (1, 1, 3), (1, 3, 1), and (3, 1, 1). Therefore, the probability of 10 5 rolling a sum of 5 or less is given by , or approximately 4.63%. 216 108 7. Since randomly selecting 5 Home Depot employees out of a total of 24 employees, or 4 require any sort of ordering, we must use combinations cashiers out of 10, does not here. The probability that exactly 4 out of the 5 randomly selected employees are C C 210 14 2,940 , cashiers is then given by the counting principle as 10 4 14 1 42, 404 42, 404 24 C5 or approximately 6.92%. CHALLENGING PROBLEMS 1. Let E1 be the event of rolling at least one ace in 4 rolls of a die and E 2 be the event of rolling at least one double aces (a.k.a. “snake eyes”) in 24 rolls of a pair of dice. Then E1C , the complementary event of E1 , consists of rolling no aces in 4 rolls of a die, and E2 C , the complementary event of E 2 , consists of rolling nodouble aces in 24 rolls of a pair of dice. We thus have the following probabilities: 5 625 671 P E1 1 P E1C 1 1 51.77%, 6 1,296 1,296 35 4 C P E 2 1 P E 2 1 49.14% . 36 and 4 Therefore, you are only slightly more likely to roll an ace in 4 rolls than “snake eyes” in 24 rolls! 2. Since ordering is not important when picking Powerball numbers, the total number of possible Powerball combinations/tickets is given by 59 C5 39 C1 5,006,386 39 195,249,054 . Here we have the following 3 winning probabilities for prizes of over $10,000: P1 P win Grand Pr ize 1 , 195,249,054 since there is only one winning Grand Prize combination (Jackpot!) P2 P win $200,000 since there are 38 ways to pick the 5 correct white balls with an incorrect red Powerball. P1 P win $10,000 38 , 195,249,054 5 C4 54 C1 195,249,054 270 , 195,249,054 since there are exactly 270 ways to pick 4 correct white balls, one incorrect white ball, and a correct red Powerball. 1 38 270 309 0.00012%, you have, therefore, 195,249,054 195,249,054 about one ten thousandth of a percentage chance of winning $10,000 or more at Powerball! Since the ordering of the cards being dealt is not important when playing poker, the total number of possible poker hands is given by 52C5 2,598,960 . Since P1 P2 P3 3. By the counting principle, the probability of drawing a one pair hand is then given by P one pair 13 4 C2 4 12 C3 3 2,598,960 13 6 64 220 1, 098, 240 42.26% , 2,598,960 2,598,960 since there are 4 C2 6 ways to select a pair from 4 cards of the same “kind” (for a “kind” of card example, 2’s, Kings, or 10’s are all “kinds” of cards), 13 ways to select for the pair, 12 C3 220 ways to select the next 3 “kinds” of distinct cards in the hand, and 43 64 ways to select one of each of those 3 cards out the 4 cards in its “kind”. By the counting principle, the probability of drawing a no pair hand (or high card hand) is then given by P no pair 13 C5 45 13 C5 4 10 45 40 2,598,960 1,302,540 50.12%, 2,598,960 C5 1, 287 ways of selecting 5 distinct “kinds” of card out of the 13 “kinds” of cards in a standard deck, 45 1,024 ways to select one of each of those 5 since there are 13 cards out the 4 cards in its “kind”, and we must subtract all 13 C5 4 5,148 possible flushes from the hand, as well as all 10 45 10, 240 straights from the hand, but we have to avoid repeating the 40 straight flushes in the hand. Therefore, you are more likely to draw a no pair (high card) hand than a one pair hand in poker. BONUS PROBLEMS 1. First, note that no ordering is required when you randomly select and write down these numbers. Secondly, if you have exactly 4 matching numbers, then you must have 3 numbers that do not match. Given all this, we get the following probability: 7 C4 13 C 1 35 286 10, 010 12.91% . 24 2. C5 77,520 77,520 If the gambler selects the fair coin, then there are four possibilities for the two flips: HH, HT, TH, and TT. If he select the two-headed coin, then there is only one possible outcome for the two flips: TT. Since there are five possible outcomes altogether, his probability of getting two heads from the fair coin is 1/5, or 20%. Note: You could also use a tree diagram for this problem.