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MAT 102 Solutions – Take-Home Exam 3 1. Applying the Counting Principle Here Yields a Sample Space Consisting Of

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MAT 102 Solutions – Take-Home Exam 3 1. Applying the Counting Principle Here Yields a Sample Space Consisting Of MAT 102 Solutions – Take-Home Exam 3 1. Applying the counting principle here yields a sample space consisting of �� = �� outcomes (e.g. �����, �����, or �����) when flipping a fair coin 5 times in a row. The event is getting at least 2 heads, so its complementary event is getting either no heads or only one head. There are exactly 6 outcomes in this complementary event: �����, �����, �����, �����, �����, ����� ) +) �� Therefore, the probability of getting at least 2 heads is 1 − = = , or 81.25%. *+ *+ �� Note: You could also use a tree diagram for this problem. 2. There are ���� = �� × �� × �� = ���� possible trifecta wagers. We use permutations for this computation since order matters for this bet. On the other ��×��×��×�� hand, there are � = = ���� possible superfecta box wagers. We use �� � �×�×�×� combinations for this computation since order doesn’t matter for this bet. The odds in favor of winning each of these wagers are, therefore, 1 in 2729 for the trifecta wager and 1 in 1364 for the superfecta box wager. Given any amount of money, you should definitely opt for the superfecta box wager as this bet is two times more likely to win than the trifecta wager. 3. There is only one way to select a pair of red Queens. Once this selection is made, ��×��×�� there are now � = = ��� ways to select the other three distinct �� � �×�×� types of cards (e.g. a Jack, a 7 and a 9) out of the remaining 12 types of cards. We use combinations here since the order of selection doesn’t matter in poker. Also, keep in mind that another black Queen could not be selected as one of the other cards since that would change the hand to a three-of-a-kind. Now, since there are two ways to select black cards for any type of card (e.g. a Jack selection could be a Jack of Spades or a Jack of Clubs), applying the counting principle � here gives a total of � × ( ����) × � = ��� × � = ���� ways to select the red pair of Queens and three other distinct black cards. ��×��×��×��×�� The sample space here consists of all � = = �, ���, ��� �� � �×�×� ×�×� possible poker hands. ���� As a result, the probability of getting this poker hand randomly dealt to you is , �,���,��� or approximately �. ����%. 4. Applying the counting principle here yields a sample space consisting of �� = ��� outcomes (e.g. (1, 4, 6) or (2, 2, 2)) when rolling three regular dice. A sum that is prime and less than 11 occurs for the following 22 outcomes: Sum of 3: (1, 1, 1) Sum of 5: (1, 1, 3), (1, 3, 1), (3, 1, 1), (1, 2, 2), (2, 1, 2), (2, 2, 1) Sum of 7: (1, 1, 5), (1, 5, 1), (5, 1, 1), (1, 2, 4), (1, 4, 2), (2, 1, 4), (2, 4, 1), (4, 1, 2), (4, 2, 1), (1, 3, 3), (3, 1, 3), (3, 3, 1), (3, 2, 2), (2, 3, 2), (2, 2, 3) �� �� Therefore, the probability of rolling a sum that is prime and less than 11 is = , or ��� ��� approximately ��. ��%. 5. Since randomly selecting 5 Home Depot employees out of a total of 22 employees, or 4 cashiers out of 9, does not require any sort of ordering, we must use combinations here. The probability that exactly 4 out of the 5 randomly selected employees are cashiers is computed using the counting principle as follows: 9 × 8 × 7 × 6 F J (13) ( A�C)( D*�D) 1 × 2 × 3 × 4 126 × 13 �� = = = ≈ �. ��% ++�E 22 × 21 × 20 × 19 × 18 26,334 ��� 1 × 2 × 3 × 4 × 5 6. There are 8 possible outcomes for the 4th term in your sequence: 57, 13.5, 12, 2.25, 9, 1.5, 0, or -0.75. (check these numbers by constructing a tree diagram). Exactly 4 of C � these numbers are integers, so the probability is = , or 50%. M � 7. By the same principle as in Problem 5, the probability that 5 out of the 7 songs you picked are the same as your friend’s is computed using the counting principle as follows: 7 × 6 × 5 × 4 × 3 13 × 12 F J F J ( N�E)( D*�+) 1 × 2 × 3 × 4 × 5 1 × 2 = +O�N 20 × 19 × 18 × 17 × 16 × 15 × 14 1 × 2 × 3 × 4 × 5 × 6 × 7 21 × 78 ��� = = ≈ �. ��% 77,520 ��, ��� 8. First, check that the family has 3 boys and 4 girls. Thus, the odds that a child selected at random from this family is a boy are 3 to 4. 9. There are 11 rolls where one of the die rolls a 5 [check this]. Only two of these outcomes � also has a 3: (5, 3) and (3, 5). So the probability sought here is . �� 10. Using the formula for conditional probabilities, we get: Pr(��� and ����) 6/20 � Pr(��� / ����) = = = = ��% Pr(����) 12/20 � Pr(��� and �����) 4/20 � Pr(��� / �����) = = = ≈ ��. �% Pr(�����) 9/20 � Therefore, it is likeliest that the randomly selected marble is odd-numbered given that it’s blue. 11. Let �� be the event of rolling at least one ace in 4 rolls of a die and �� be the event of rolling double aces (a.k.a. “snake eyes”) at least once in 24 rolls of a pair of dice. Then, �aaa�a, the complementary event of �D, consists of rolling no aces in 4 rolls of a die, and �aaa�a, the complementary event of �+, consists of rolling no double aces in 24 rolls of a pair of dice. The probabilities of �D and �+ are computed as follows: E C )+E ��� Pr(� ) = � − Prc� d = � − F J = 1 − = ≈ ��. ��% D D ) D+A) ���� *E +C Pr(� ) = � − Prc� d = � − F J ≈ ��. ��% + + *) Therefore, you are only slightly more likely to roll an ace in 4 rolls than “snake eyes” in 24 rolls! 12. Let � be the event “You select all 5 lucky numbers” and � be the event “You select exactly 4 lucky numbers.” Then, since these two events are mutually exclusive the probability of selecting either 4 or 5 lucky numbers is given by �(� �� �) = �(�) + �(�). ��×��×��×��×�� In both probabilities, the sample space consists of all � = = �� � �×�×� ×�×� ��, ���, ��� possible ways to select 5 random numbers (without caring about the order) out of the total 80 numbers. Now, there are ���� = ��, ��� ways of selecting 5 “lucky” numbers out of 20. Here we use combinations again since the order of selection doesn’t matter. On the other hand there are ( ����)( ����) = ���� × �� = ���, ��� ways to select exactly 4 “lucky” numbers and 1 “unlucky” number. As a result, we get DE,EOC +AO,NOO ���,��� �(� �� �) = �(�) + �(�) = + = ≈ �. ��% +C,OCO,OD) +C,OCO,OD) ��,���,��� 13. There are 44 different ways to place the balls so that no ball label matches its bin label. To check this result, do a systematic alphabetical listing of all the possibilities: start with ball B in Box A and check all possibilities resulting from this first ball placement: BADEC, BAECD, BCAED, BCDEA, etc. Then proceed in the same manner placing Ball C in Box A, and then Ball D in Box A, and finally Ball E in Box A. Keep listing all possibilities until they are exhausted. There are, in total, ��� = �! = � × � × � × � × � = ��� ways to arrange the 5 balls in the 5 bins. We use permutations for this computation since order matters when placing the balls in the bins. As a result, the odds in favor of blindly placing all 5 balls bins so that no ball label �� matches its bin label are 44 to 76, or 11to 19. This is a probability of , or �� approximately 36.7%. 14. Let � denote the number of black socks in the drawer. Then the probability that 2 socks l lnD selected at random are black is given by ∙ . This implies the following: DO A � � − 1 1 ∙ = ⇒ �(� − 1) = 30 ⇒ � = 6. 10 9 3 So there are 6 black socks and 4 white socks in the drawer. The probability that 2 socks selected at random is not a pair is then computed as follows: D C * D + � 1 − − F ∙ J = 1 − − = . * DO A * DE �� .
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