course Mth 152 GxSՎϑassignment #006
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13:17:20 Query 12.1.6 8 girls 5 boys What is the probability that the first chosen is a girl?
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RESPONSE --> the probability would be 8 out of thirteen or 8/13
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13:17:25 ** Assuming the choice is completely random there are 13 possible choices, 8 of which are female so we have P(female) = 8 / 13 = .6154, approx. **
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RESPONSE --> o.k.
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13:20:45 Query 12.1.12 3 fair coins: Probability and odds of 3 Heads.
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RESPONSE --> hhh, hht htt hth ttt 1 out of 8 tth 1 to 7 tht thh
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13:20:52 ** There are 8 equally likely possible outcomes when flipping 3 fair coins. You can list them: hhh, hht, hth, htt, thh, tht, tth, ttt. Or you can use the fact that there are 2 possibilities on each flip, therefore 2*2*2 = 2^3 = 8 possible outcomes. Only one of these outcomes, hhh, consists of 3 heads. The probability is therefore P(3 heads) = # of outcomes favorable/total number of possible outcomes = 1 / 8. The odds in favor of three heads are Odds ( 3 heads ) = # favorable to # unfavorable = 1 to 7. **
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RESPONSE --> 0.k.
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13:23:25 Query 12.1.20 P(pink) from two pink parents (Rr and Rr) What is the probability of a pink offspring.
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RESPONSE --> R r R RR Rr two out of the four would be pink 2/4=1/2 r rR rr
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13:23:31 ** The genes R and r stand for the red and white genes. A pink offspring is either Rr or rR. RR will be red, rr white. R r R RR Rr r rR rr shows that {RR, Rr, rR, rr} is the set of equally likely outcomes. We season two of the four possible outcomes, rR and Rr, will be pink. So the probability of pink offspring is 2/4 = 1 / 2. **
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RESPONSE --> o.k.
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13:24:47 Query 12.1.33 cystic fibrosis in 1 of 2K cauc, 1 in 250k noncauc What is the empirical probability, to 6 places, that a randomly chosen non-Caucasian newborn will have cystic fibrosis?
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RESPONSE --> the chance would be 1 in 250,000
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13:24:53 ** There is 1 chance in 250,000 so the probability is 1 / 250,000 = 4 * 10^-6, or .000004. **
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RESPONSE --> o.k.
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13:28:45 12.1.40 Cc genes carrier, cc has disease; 2 carriers first child has disease **** What is the probability that the first child has the disease?
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RESPONSE --> C c 1 in four chance C CC Cc c cC cc
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13:28:49 ** If cc has the disease, then the probability that the first child will have the disease is 1/4. **
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RESPONSE --> o.k.
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13:29:36 What is the sample space for this problem?
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RESPONSE --> CC,Cc, cC, cc
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13:29:41 ** The sample space is {CC, Cc. cC, cc}. **
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RESPONSE --> o.k.
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13:34:01 12.1.60 36 in class, 3 chosen **** What is the probability that the choice will be the given three people in any order?
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RESPONSE --> 36 in class, 3 chosen 36/3=12
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13:34:12 ** There are P(36,3) possible ordered choices of 3 people out of the 36. P(36,3) = 36! / (36-3)! = 36! / 33! = (36*35*34*33*32*31*...*1) / (33*32*31*...*1) = 36*35*34=40,000 or so. The probability of any given choice is therefore 1 / P(36,3) = 1/40,000 = .000025, approx.. For any given set of three people there are six possible orders in which they can be chosen. So the probability of the three given people, in any order, is 6 * probability of a given order = 6 / P(36,3) = 6/40,000 = .00015. Alternatively we can say that we are choosing 3 of 36 people without regard for order, so there are C(36,3) possibilities and the probability of any one of them is 1 / C(36,3). **
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RESPONSE --> o.k.
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13:40:14 Query 12.1.75 digits 1, 2, ..., 5 rand arranged; prob even, prob digits 1 and 5 even What is the probability that the resulting number is even and how did you obtain your answer?
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RESPONSE --> 1,2,3,4,5 ( 2,2 ) 2 choices that the number is even (5,2) choices that the numeber will be even 5x4= 20= 2 divided by 20 2/20 = 1/10
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13:40:26 ** The number will be even if it ends in 2 or 4. There are 5 possible ending numbers. So 2 of the 5 possible ending numbers are even and the probability of an even number is 2/5. We analyze in two ways the number of ways to choose a number with digits 1 and 5 even. First way: There are 5! = 120 possible arrangements of the 5 digits. There are only two possible even digits, from which we will choose digit 1 and digit 5. The order of our choice certainly matters, since a different choice will give us a different 5-digit number. So we are choosing 2 digits from a set of 2 digits, where order matters. We therefore have P(2, 2) = 2*1 / 0! = 2 ways to choose these digits. The remaining 3 digits will comprise digits 2, 3 and 4. We are therefore choosing 3 digits from a set of 3, in order. There are P(3, 3) = 3*2*1/0! = 6 ways to do so. To obtain our number we can choose digits 1 and 5, then digits 2, 3 and 4. There are P(2, 2) * P(3, 3) = 2 * 6 = 12 ways to do this. So the probability that digits 1 and 5 are even is 12 / 120 = 1/10. Second way: A simpler solution looks at just the possibilities for digits 1 and 5. There are P(2, 2) = 2 choices for which these digits are even, and P(5, 2) = 20 total choices for these two digits. The probability that both will be even is therefore 2/20 = 1/10, the same as before. **
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RESPONSE --> o.k.
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