course mth152 I have most of my assignments finished but they are on paper, I will try to enter these into the computer this weekend.Dave ?·c·???q???·????`??·assignment #008
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11:34:09 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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RESPONSE -->
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11:35:35 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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RESPONSE --> The first choice is independent. However, as far as the second choice it depends on whether or not the first caused the remaining members to be less Republican or more Republican.
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11:35:44 ** 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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RESPONSE --> Yes, more or less what I was trying to get at
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11:39:09 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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RESPONSE --> As for females, there are a total of 42... out of these, 14 are driven by money, and 13 are focused on creativity. So 14+13=27, out of 42 there are 27 total interested... 27/42= .643
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11:39:20 ** 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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RESPONSE --> Hmmm... my question in the book looks a little different
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11:44:29 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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RESPONSE --> The probaility of this is 13/51, as a total of 13 club cards are still in the remaining deck of 51 cards (minus the diamond). If you wanted the probability that a diamond would be first, however, and a club second I may need a little help on that... Actually, I have a question.... how would that work? the probability of a diamond first and a club second?
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11:44:37 ** 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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RESPONSE --> Ah, I see.
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11:45:47 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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RESPONSE --> The probability of getting a diamond out of the total of red cards (26) is 13/26 or 1/2=.5000
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11:45:53 ** 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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RESPONSE --> Exactly
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11:48:58 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 --> As there is no binding connection between the three sales that I can see these are all independent in regards to their probailities... As such, I would calculate them as P(.8,3) or .8*.8*.8=.512
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11:49:04 ** 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 --> Exactly
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11:50:45 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 --> I had much the same outcome for this assignment... as .95 is the probability of none of the critical cloud movements, I would use P (.95, 5) .95*.95*.95*.95*.95=.774
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11:50:50 ** 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 --> Exactly
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11:52:09 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 --> probility of heads is 52/100 or 26/50 or 13/25 probility of tails is 48/100 or 24/50 or 17/25
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11:52:21 ** 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 --> Okay, I figured this wrong I think
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11:53:40 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 --> As there was no rain before the first of the four days the probability of the first day is .3, after that the probility for the rest is .8 (due to previous rain)
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11:53:45 ** 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 --> Yes
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11:54:13 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 --> I think I had the same problem?
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11:54:34 ** 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 --> Oh, sorry... I looked at the wrong one in my book I believe for the previous problem
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11:54:43 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 --> i see
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11:54:51 ** 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 --> okay
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11:55:02 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> I believe I did a wee bit better on this one
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