course Mth 152 I will send the first part of this assignment tomorrow. I worked on it at school and Daddy got sick and I had to leave and didn't have a chance to send it to you. Daddy has been doing good but he got sick today with a headache and had a lot of fluid so bad that it scared us so we took him to the ER but they really didn't tell us anything. They felt that the headache is still some of the after effect of the cyberknife. But we did get some great news, while we where there the doctor looked at Daddy's scans he had done this past Friday to see how the tumor has done with the cyberknife treatment. We were blessed with the news that the tumor has srunk. I thought I would let you and Bridgette know. ?··??????R??????
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18:47:19 Query 12.3.6 two members chosen for committee, Republican or no. Are the two choices independent or dependent and why?
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18:47:22 ** The choice of the second is influenced by the first choice. If a Republican is chosen on the first choice, then there are fewer Republicans available for the second choice and the probability of getting a Republican on the second choice is lower than if a Republican had not been chosen first. COMMON ERROR: they are independant because they were randomly selected....if it is random then one did not depend on the other. EXPLANATION: The selection was indeed random, but the makeup of the remaining group available on the second choice depends on the first choice. **
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18:47:25 Query 12.3.12 table of motivations by male, female What is the probability that an individual will be primarily motivated by money or creativity given that the individual is female?
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18:47:28 ** There are 13 primarily motivated by money and 14 by creativity. Each person can have only one primary motivation so there is no overlap between these two groups. There are thus 13 + 14 = 27 motivated by money or creativity, out of a total of 66 women, which gives probability 27 / 66 = 9/22 = .41 approx.. If M is the set motivated by money, C the set motivated by creativity and S the entire sample space then we have p ( M or C) = p(M) + p(C) - p(M and C) or in set notation p(M U C) = p(M) + p(C ) - p(M ^ C), with U and ^ standing for set union and intersection. Since M ^ C is empty, p(M ^ C) = 0. p(M) = n(M) / n(S) = 13 / 66 and p(C) = n(C) / n(S) = 14/66 so we have p (M U C) = 13/66 + 14/66 = 27/66 = 9/22 = .41 approx. **
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18:47:35 Query 12.3.24 prob of club 2d given diamond first What is the desired probability and how did you obtain it?
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18:47:39 ** The probability is 13 / 51. When the second card is chosen there are 13 clubs still left, out of 51 remaining cards. This can also be calculated using P(A|B) = P(A^B) / P(B). The probability of getting a diamond on the first card and a club on the second is 13/52 * 13/51. The probability of getting a diamond on the first card is 13/52. So the probability of a club on the second given a diamond on the first is (13 / 52 * 13 / 51) / (13/52) = 13/51. **
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18:47:44 Query 12.3.32 prob of diamond given red What is the probability of getting a diamond given that the card is red?
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18:47:48 ** Of the 26 red cards, 13 are diamonds. So the probability of a diamond, given red, is 13/26 = 1/2. **
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18:53:03 Query 12.3.36 P(sale > $100) = .8; prob that first three sales all >$100 What is the probability that the first three sales are all for > $100 and how did you obtain your result?
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RESPONSE --> .8*.8*.8 = .51 since there is an .8 probability of a sale of more than $100, then 3 sales would be .8*.8=.64*.8 = .51
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18:53:34 ** the first sale has to be > $100, AND the second sale has to be > $100, AND the third sale has to be > $100. The events are independent. So the total probability, by the fundamental counting principle, is .8 * .8 * .8 = .512. **
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RESPONSE --> That is what I got too. I multipled the probablity of each sell three times.
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19:47:23 Query 12.3.42 P(critical direction) = .05; prob that none of the 5 scheduled days for launches has cloud movement in critical direction
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RESPONSE --> .05*0 = 0
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19:48:15 ** On a given day cloud movement is not in critical direction with probability 1 - .05 = .95. this has to occur on the first day, then it has to occur on the second day, then on the third, then on the fourth, then on the fifth. These events are considered independent so the probability is .95 * .95 * .95 * .95 * .95 = .774 approx. (use your calculator to get the accurate answer). In order for none of the five days to have cloud movement in the critical direction, each of the five days must not have movement in the critical direction. The probability that the movement will not the in the critical direction for each of the days is .95. The probability that this will happen on every one of the five days is therefore .95 * .95 * .95 * .95 * .95 = .774, approx. **
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RESPONSE --> I did not do the 1-.05 to get .95 so I got the wrong answer. I mutiplied it by zero.
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19:53:23 Query 12.3.54 probability of heads .52, tails .48; P(ht) What is the probability of head then tails?
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RESPONSE --> .52*.48 = .2496
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19:53:57 ** There is a .52 probability of getting heads, then there is a .48 probability a getting tails. The two events have to happen consecutively. By the Fundamental Counting Principle there is thus a probability of .52 * .48 = .2496 of getting Heads then Tails. **
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RESPONSE --> I mutiplied the probablity of heads with the probablity of tails to get the answer.
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20:21:51 Query 12.3.66 for given day P(rain)=.5, P(rain | rain day before) = .8, P(rain | no rain day before ) = .3. find P(rain on 3 consecutive days). For first 4 days in November what is the probability that it will rain on all four days given Oct 31 is clear?
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RESPONSE --> .3*.5*.5*.8 = .060
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20:22:57 ** The probability of rain on the first of the four days is .3, since it is given that there was no rain on the previous day. The probability of rain on each of the following for days is .8, since on each of these days it rained the day before. The probability of rain on all four days is therefore .3 * .8 * .8 * .8 = .154. ANOTHER WAY OF SAYING IT: Oct 31 was clear so the probability of rain on the first day is .3. If it rained on the first day of the month then there is a probability of .8 that it rains on the second day. If it rained on the second day of the month then there is a probability of .8 that it rains on the third day. If it rained on the third day of the month then there is a probability of .8 that it rains on the fourth day. So the probability of rain on all 4 days is .3 * .8 * .8 * .8 = .154 **
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RESPONSE --> I mutiplied by .5 instead of .8 to end the rain days.
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20:25:19 Query 12.3.66 for given day P(rain)=.5, P(rain | rain day before) = .8, P(rain | no rain day before ) = .3. What is P(rain on 3 consecutive days).
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RESPONSE --> 1st day rain, 2nd day rain/rain day before, 3rd day rain/rain day before .5*.8*.8* = .320
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20:25:45 ** To get rain on 3 consecutive days requires rain on the first day, which happens with probability .5; then rain on the second day given that there was rain on the first day, which is .8, then rain on the third day, given that there was rain on the previous day; this third probability is also .8. The probability of the 3 events all happening (rain of 1st day AND rain on the second day AND rain on the third day) is therefore .5 * .8 * .8 = .32. **
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RESPONSE --> I got the days straight this time. I think typing the days out helped.
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20:31:52 QUESTION ON PROBLEM 33: Please explain Problem 33 of 12.3. It reads: If one number is chosen randomly from the intergers 1 throught 10, the probability of getting a number that is odd and prime, by the general multiplication rule is P(odd) * P(prime/odd) = 5/10 * 3/5 = 3/10 My question is how did we get three prime numbers out of 1 through 10? I assumed there were 4 of them (2, 3, 5, and 7).
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RESPONSE --> 2 maybe prime but it even instead of prime.
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20:32:55 ** ONE ANSWER: The sample space is reduced to odd numbers, and 2 is not odd. So the set within the restricted sample space {1, 3, 5, 7, 9} should just be {3, 5, 7}. ANOTHER ANSWER: If we don't use the restricted sample space then we have P(prime | odd ) = P(prme) * P(odd | prime). We find P(prime) and P(odd | prime). P(prime) = 4 / 10, since there are 4 primes between 1 and 10. Within the unrestricted sample space P(odd | prime) is 3 / 4 since of the primes 2, 3, 5, 7 only three are odd. }Thus when you multiply P(prme) * P(odd | prime) you get 4/10 * 3/4 = 3/10, just as before. **
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RESPONSE --> That is what I thought, since 2 was even instead of odd and eventhough it is prime, it would not be included.
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20:33:02 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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