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Discrete Mathematics Math 6A

Discrete Mathematics Math 6A. Homework 5 Solution. 5.1-6 There are 16 cards that qualify as being an ace or a heart, so The prob. that a card selected from a deck is an ace or a heart = 16/52

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Discrete Mathematics Math 6A

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  1. Discrete MathematicsMath 6A Homework 5 Solution

  2. 5.1-6 There are 16 cards that qualify as being an ace or a heart, so The prob. that a card selected from a deck is an ace or a heart = 16/52 5.1-15 We have to choose five cards. That is, two pairs and another card that is not the same number with two pairs. There are four kinds of card and each kind has 13 cards. So the total number of cards is 52(=13*4) For example, we can choose 1 heart + 1 diamond + 2 heart + 2 diamond + another number that is not the same with 1 or 2 among four kinds. There are C(13,2)*C(4,2)*C(4,2)*(52-8) = 78*6*6*44=123,552 different hands.Since each hand is equally likely, and there are C(52,5) = 2,598,960 different hands. Therefore, the probability of holding two pairs is 123,552/ 2,598,960 5.1-18 A "straight flush" means choosing a consecutive five cards of the same kind among the four kinds. For example, we can choose 12345, 23456,..., 10,J,Q,K,A that is 10 ways for choosing the straight flush. So there are 10*4 straight flushes. Therefore, the prob. to choose the straight flush is 40/C(52,5) = 1/64974 Note: 10,J,Q,K,A is actually "Royal Straight Flush", but we count this as a straight flush for this problem. Note: From the J,Q,K,A,1 or Q,K,A,1,2 or etc are NOT the straight flush !

  3. 5.1-27 First, we need to select exactly one of the correct six integers among the n numbers and the order does not matter. There are C(n,6) possible choices to pick the correct six integers and we need to select one of C(n,6). Second, there are n-6 incorrect integers and we need to select rest of five incorrect integers among the C(n-6,5). Third, there are six ways to choose exactly one correct integers among the six correct integers. So the prob. of selecting exactly one of the correct six integers is 6{C(n-6,5)/C(n,6)} • if n=40, then prob. is 6{C(40-6,5)/C(40,6)} = 6{C(34,5)/C(40,6)}= 139128/319865 • if n=48, then prob. is 6{C(48-6,5)/C(48,6)} = 6{C(42,5)/C(48,6)}= 212667/511313 • if n=56, then prob. is 6{C(56-6,5)/C(56,6)} = 6{C(50,5)/C(56,6)}= 151340/386529 • if n=64, then prob. is 6{C(64-6,5)/C(64,6)} = 6{C(58,5)/C(64,6)}= 163647/446276 • 5.1-32 100 people for a contestant and different winners are selected randomly. • First , there are three ways for one of K, J, and P to be picked to win first prize, two ways for one of the others to win second prize, and one way for the third to win third prize. Second, with the same logic above, there are 100 ways to pick as the first prize, and 99 ways to pick as the second prize and 98 ways to pick as the third prize. Therefore, prob. is (3*2*1)/(100*99*98) = 1/161700

  4. 5.1-37 First, there are four ways to get a total of 9 when two dice are rolled: (6,3), (5,4), (4,5), (3,6) So, the prob. of getting of 9 is 4/(6^2). Second, there are 26 ways to get a total of 9 when three dice are rolled: (6,2,1), (6,1,2) and (5,3,1), (5,2,2), (5,1,3) and (4,4,1), (4,3,2), (4,2,3), (4,1,4) and (3,5,1),(3,4,2),(3,3,3), (3,2,4),(3,1,5) and (2,6,1),(2,5,2),(2,4,3),(2,2,5),(2,5,2),(2,3,4),(2,1,6) and (1,6,2),(1,5,3),(1,4,4),(1,3,5),(1,2,6) So the prob. getting of 9 is 26/(6^3). Therefore, three dice is more likely 5.1-40 First, you had a ¼ chance of winning with your original selection. Second, Your chance of winning after switching is half of ¾, that is 3/8

  5. 5.2-5 First, probabilities for 4 appears on the first die is 2/7 and 3 appears on the second dies is 2/7 and other outcomes for each die appear is 1/7. Second, there are six ways to roll a sum of 7: (1,6),(2,5),(3,4),(4,3),(5,2),(6,1) where (i,j) means rolling i on the first die and j on the second. • p((1,6)) = 1/7*1/7=1/49, p((2,5)) = 1/7*1/7=1/49, p((3,4)) = 1/7*1/7=1/49, • p((4,3)) = 2/7*2/7=4/49, p((5,2)) = 1/7*1/7=1/49, p((6,1)) = 1/7*1/7=1/49 • and sum of above probs. is 9/49 • 5.2-7 {1,2,3,4} • We can simply think that there are only 1 and 4 and we don't have to think about 3 and 4. So in this case, 14 or 41 so the Prob.(1 precedes 4) = ½ • With the same logic with a), so the Prob.(4 precedes 1) = ½ • Prob.(4 precedes 1 and 2) = 1/3, because the same concept with a) and b), so just think about the permutation of three numbers that is, 412 and just put the 4 in the first position of the three possible positions • Prob.(4 preceds1,2 and 3) = ¼, just put the 4 in the first position in possible four positions • Prob.(4 precedes 3 AND 2 precedes 1) = Prob.(4 precedes 3)* Prob.(2 precedes 1)=1/2*1/2=1/4 because 4 and 2 are not related with each other, that is independent.

  6. 5.2-12 p(E)=0.8 and p(F)=0.5 • Cleary p(E F) >= p(E) = 0.8. Also, p(E F) <=1. If we apply Th2 from section 5.1, we can rewrite this as p(E)+p(F)-p(EF) <= 1, or 0.8 + 0.6 – p(EF) <=1. Solving for p(EF) gives p(EF) >= 0.4 • 5.2-18 We assume for simplicity here that births are independent and the probability of a birth in each day is 1/7 • The prob. that the second person has the same birth day-of-the-week as the first person is 1/7, that is p(two people were born on the same day of the week) = 1/7 • The prob. that all the birth days-of-the-week are different is p_n= 6/7*5/7*...(8-n)/7 since each person after the first must have a different birth day-of-the-week from all the previous people in the group. Note that if n>=8, then p_n=0 since the seventh fraction is 0 (this also follows from the pigeonhole principle). The prob. that at least two are born on the same day of the week is therefore 1-(p_n) • We compute 1-(p_n) for n=2,3,... and find that the first time this exceeds ½ is when n=4(if n=4 then 1-(p_n) >= ½), so that is our answer. With four people, the prob. that at least two will share a birth day-of-the-week is 1-(6/7*5/7*4/7) = 223/343, or about 65%

  7. 5.2-23 By def3. The conditional prob. of E given F is p(E|F) = p(EF)/p(F) • Flipping the coins five times: number of trials • p(F)= 16, the first flip came up heads • (10000~11111 among 2^5=32: 00000 ~11111, if 1=head and 0=tail) • p(E)= 4, excatly four heads appear • p(E|F) = p(EF)/p(F) = 4/16 = ¼ • 5.2-31 The prob. that a family with five children does not have a boy, if the sexes of children are independent and if • If a boy and a girl are equally likely, then the prob. is (½)^5 = 1/32 • If the prob. of a boy is 0.51, then the prob. of a girl is 0.49, so (0.49)^5  0.028 • If the prob. that the ith child is a boy is 0.51-(i/100), then • p(1) = 0.51-1/100, p(2) = 0.51-2/100 ,..., p(5) = 0.51-5/100 • 1-p(1) * 1-p(2) *...* 1-p(5) = 0.50*0.51*0.52*0.53*0.54  0.038

  8. 5.2-38 Two dice rolling, the observer informs at least one of the dice came up six. • prob. of sum is seven, given the inform. provided by the honest observer. • p(E) is sum 7 • p(F) is at least one dice came up 6 and p(F) = (6^2 – 5^2) /6^2 = 11/36 • p(EF) = 2/36 (1,6) and (6,1) • Therefore, the answer p(E|F) = (2/36)/(11/36) = 2/11 • (b) Suppose that the honest observer tells us that at least one die came up five. • Then the analysis is exactly the same as in part (a), so the answer is again 2/11 • p(E) is sum 7 • p(F) is at least one dice came up 5 and p(F) = (6^2 – 5^2) /6^2 = 11/36 • p(EF) = 2/36 (2,5) and (5,2) • Therefore, the answer p(E|F) = (2/36)/(11/36) = 2/11

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