#$&* course Mth 152 October 10 - 8:15pmHad trouble reading questions in query. Words were not whole and jumbled and it led to confusion about what was being asked. Where are these questions located again??? I thought they were coming from the questions in the book but I cannot locate them?? Is it 12.1.16 ... does that mean chapter 12 section 1 question 16???
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Given Solution: `a ** Assuming the choice is completely random there are 13 possible choices, 8 of which are female so we have P(female) = 8 / 13 = .6154, approx. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): okay ------------------------------------------------ Self-critique Rating: okay ********************************************* question: `q Query 12.1.12 3 fair coins: Probability and odds of 3 Heads. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 3 fair coins. 3 heads and 3 tails. 2 sides to each coin. If we flip the coins there is a possibility of 8 outcomes/2 per flip of one coin = 2*2*2 = 8. There is only one possibility that all three coins could have three heads so it’s a possibility of 1/8 that it’s going to happen. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThere 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. ** STUDENT QUESTION I don’t understand where to 7 came from. I got there are 8 possibilities. INSTRUCTOR RESPONSE Odds = # favorable to # unfavorable. In this case there we have 1 favorable and 7 unfavorable outcomes. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Ahh, I see. Yes. Okay. Favorable = 1/8. Unfavorable = 1/7. ------------------------------------------------ Self-critique Rating: okay ------------------------------------------------ Self-critique Rating: okay ********************************************* question: `q Query 12.1.20 P(pink) from two pink parents (Rr and Rr) What is the probability of a pink offspring. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: With have two pink parents = Rr and Rr. RR stands for red and rr = white. We are trying to predict the color of the their offspring. For the offspring to be pink it has to be either Rr or rR. We use a Punnett square. R r R RR Rr r rR rr the probability of a pink baby flower is ½. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a The genes R and r stand for the red and white genes. A pink offspring is either Rr or rR. RR will be red, rr white. R r R RR Rr r rR rr shows that {RR, Rr, rR, rr} is the set of equally likely outcomes. We season two of the four possible outcomes, rR and Rr, will be pink. So the probability of pink offspring is 2/4 = 1 / 2. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): okay ------------------------------------------------ Self-critique Rating: okay ********************************************* question: `q Query 12.1.33 cystic fibrosis in 1 of 200,000 K cauc, 1 in 250,000 k non Caucasian What is the empirical probability, to 6 places, that a randomly chosen non-Caucasian newborn will have cystic fibrosis? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I do not understand this question. I’d say 1 in 250,000 for non Caucasian but to 6 places….. what does that mean??? confidence rating #$&*: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThere is 1 chance in 250,000 so the probability is 1 / 250,000 = 4 * 10^-6, or .000004. ** question 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I can’t find this problem and It doesn’t read very well I don’t know if I’m using the correct information??? There are 2 carriers = Cc and cc??? When you use the Punnett Square, there will be four opportunities. But since we only have one dominant trait in the one carrier and the rest are small c’s the square turns out like this. C c c cC cc c cC cc confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aIf cc has the disease, then the probability that the first child will have the disease is 1/4. ** My answer doesn’t match that at all??? What is the sample space for this problem? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: cC, cc, cC, cc confidence rating #$&*: 0 - I know this is wrong. Can you explain what I did wrong? I didn’t understand the information given in the first question above about the carriers??? ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe sample space is {CC, Cc. cC, cc}. ** 12.1.61 & 60 & 36 in class, 3 chosen **** What is the probability that the choice will be the given three people in any order? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 36 in class and 3 are chosen - what is the probability that the choice will be the given three people in any order? The permutation formula is P ( 36,3) P (36,3) 36! /(36 - 3!) 36! /33! 36*35*34 42,840 To determine the probability would be 1/42,840 for each choice or 1/ P(36,3) confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThere are P(36,3) possible ordered choices of 3 people out of the 36. P(36,3) = 36! / (36-3)! = 36! / 33! = (36*35*34*33*32*31*...*1) / (33*32*31*...*1) = 36*35*34=40,000 or so. The probability of any given choice is therefore 1 / P(36,3) = 1/40,000 = .000025, approx.. For any given set of three people there are six possible orders in which they can be chosen. So the probability of the three given people, in any order, is 6 * probability of a given order = 6 / P(36,3) = 6/40,000 = .00015. Alternatively we can say that we are choosing 3 of 36 people without regard for order, so there are C(36,3) possibilities and the probability of any one of them is 1 / C(36,3). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I guess I needed to round down to 40,000 to make the percentage and decimals easier to figure. ------------------------------------------------ Self-critique Rating: 2
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Given Solution: `aThere is 1 chance in 250,000 so the probability is 1 / 250,000 = 4 * 10^-6, or .000004. ** question 12.1.40 Cc genes carrier, cc has disease; 2 carriers first child has disease **** What is the probability that the first child has the disease? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I can’t find this problem and It doesn’t read very well I don’t know if I’m using the correct information??? There are 2 carriers = Cc and cc??? When you use the Punnett Square, there will be four opportunities. But since we only have one dominant trait in the one carrier and the rest are small c’s the square turns out like this. C c c cC cc c cC cc confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aIf cc has the disease, then the probability that the first child will have the disease is 1/4. ** My answer doesn’t match that at all??? What is the sample space for this problem? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: cC, cc, cC, cc confidence rating #$&*: 0 - I know this is wrong. Can you explain what I did wrong? I didn’t understand the information given in the first question above about the carriers??? ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe sample space is {CC, Cc. cC, cc}. ** 12.1.61 & 60 & 36 in class, 3 chosen **** What is the probability that the choice will be the given three people in any order? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 36 in class and 3 are chosen - what is the probability that the choice will be the given three people in any order? The permutation formula is P ( 36,3) P (36,3) 36! /(36 - 3!) 36! /33! 36*35*34 42,840 To determine the probability would be 1/42,840 for each choice or 1/ P(36,3) confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThere 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). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I guess I needed to round down to 40,000 to make the percentage and decimals easier to figure. ------------------------------------------------ Self-critique Rating: 2
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Given Solution: `aThere is 1 chance in 250,000 so the probability is 1 / 250,000 = 4 * 10^-6, or .000004. ** question 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I can’t find this problem and It doesn’t read very well I don’t know if I’m using the correct information??? There are 2 carriers = Cc and cc??? When you use the Punnett Square, there will be four opportunities. But since we only have one dominant trait in the one carrier and the rest are small c’s the square turns out like this. C c c cC cc c cC cc confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aIf cc has the disease, then the probability that the first child will have the disease is 1/4. ** My answer doesn’t match that at all??? What is the sample space for this problem? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: cC, cc, cC, cc confidence rating #$&*: 0 - I know this is wrong. Can you explain what I did wrong? I didn’t understand the information given in the first question above about the carriers??? ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe sample space is {CC, Cc. cC, cc}. ** 12.1.61 & 60 & 36 in class, 3 chosen **** What is the probability that the choice will be the given three people in any order? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 36 in class and 3 are chosen - what is the probability that the choice will be the given three people in any order? The permutation formula is P ( 36,3) P (36,3) 36! /(36 - 3!) 36! /33! 36*35*34 42,840 To determine the probability would be 1/42,840 for each choice or 1/ P(36,3) confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThere 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). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I guess I needed to round down to 40,000 to make the percentage and decimals easier to figure. ------------------------------------------------ Self-critique Rating: 2