course Mth 152 㼯izUw[_Ȋjqyassignment #006 QinwϏ̧I assignment #007 cʓ齯̀q}li Liberal Arts Mathematics II 10-11-2006
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19:35:36 12.2.6 single die, p(odd or <5). What is the probability of getting an odd result or a result < 5?
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RESPONSE --> odd number 3/6=0.5 <5 4/6=0.67
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19:37:12 ** there are 3 possible odd outcomes and four outcomes less that 5 which would add up to 7 outcomes, except that 2 of the outcomes < 5 are alrealdy odd and won't be counted. Thus the number of outcomes which are odd or less that 5 is 3 + 4 - 2 = 5 (this expresses the rule that n(A U B) = n(A) + n(B) - n(A ^ B), where U and ^ stand for union and intersection, respectively ). Thus the probability is 5/6. In terms of the specific sample space: The sample space for the experiment is {1, 2, 3, 4, 5, 6}. Success corresponds to events in the subset {1, 2, 3, 4, 5}. There are 6 elements in the sample space, 5 in the subset consisting of successful outcomes. Thus the probability is 5/6. **
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RESPONSE --> ok
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19:38:55 Query 12.2.15 drawing neither heart nor 7 from full deck
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RESPONSE --> probability- 9/13=0.69 odds- 9 to 4
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19:39:08 ** The sample space consists of the 52 cards in a full deck. There are 39 cards that aren't hearts, four 7's but only three aren't hearts so there are 36 cards that aren't hearts or seven. The probability is therefore 36/52 = 9/13. The odds in favor of the event are 16 to 36 (number favorable to number unfavorable), which in reduced form is 4 to 9. **
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RESPONSE --> ok
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19:42:41 12.2.24 prob of black flush or two pairs
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RESPONSE --> black flush=5108/2598960=0.001965 two pairs=123552/2598960=0.047539
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19:43:27 ** There are C(13,5) = 1287 ways to get a flush in a given suit--gotta choose the 5 cards from the 13 cards in that suit. There are two black suits so there are 2 * 1287 = 2574 possible black flushes. As the text tells you there are 123,552 ways to get two pairs. You can incidentally get this as 13 * C(4, 2) * 12 * C(4, 2) * C(44, 1) / 2 (2 of the 4 cards in any of the 13 denominations, then 2 of the 4 cards in any of the remaining 12 denominations, divide by 2 because the two denominations could occur in any order, then 1 of the 44 remaining cards not in either of the two denominations. There is no way that a hand can be both a black flush and two pairs, so there is no overlap to worry about (i.e., n(A and B) = 0 so n(A or B) = n(A) + n(B) - n(A and B) = n(A) + n(B) ). Thus there are 123,552 + 2574 = 126,126 ways to get one or the other. The probability is therefore 126,126 / 2,598,960 = .0485, approx. **
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RESPONSE --> ok
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19:45:30 12.2.33 x is sum of 2-digit numbers from {1, 2, ..., 5}; prob dist for random vbl x
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RESPONSE --> x P(x) 3 0.1 4 0.1 5 0.2 6 0.2 7 0.2 8 0.1 9 0.1
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19:45:41 ** If 2 different numbers are chosen from the set (1, 2, 3, 4, 5} then the sum 3 can appear only as 1+2. 4 can appear only as 1+3, assuming numbers can't be repeated (so, for example, 2+2 is not allowed). 5 can occur as 1+4 or as 2+3. 6 can occur as 1+5 or as 2+3. 7 can occur as 2+5 or as 1+6. 8 can occur only as 3+5. 9 can occur only as 4+5. Of the 10 possible combinations, the sums 3, 4, 8 and 9 can occur only once each, so each has probability .1. The sums 5, 6 and 7 can occur 2 times each, so each has probability .2. The possible sums are as indicated in the table below. 1 2 3 4 5 1 3 4 5 6 2 5 6 7 3 7 8 4 9 This assumes selection without replacement. There are C(5, 2) = 10 possible outcomes, as can be verified by counting the outcomes in the table. 3, 4, 8 and 9 appear once each as outcomes, so each has probability 1/10. 5, 6 and 7 appear twice each as outcomes, so each has probability 2/10. x p(x) 3 .1 4 .1 5 .2 6 .2 7 .2 8 .1 9 .1 **
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RESPONSE --> ok
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19:52:12 Query 12.2.36 n(A)=a, n(S) = s; P(A')=? What is the P(A')?
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RESPONSE --> It is the probability that an event will occur within sample space S
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19:52:32 ** A' is everything that is not in A. There are a ways A can happen, and s possibilities in the sample space S, so there are s - a ways A' can happen. So of the s possibilities, s-a are in A'. Thus the probability of A' is P(A') = n(A') / n(S) = (s - a) / s. **
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RESPONSE --> ok
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19:55:50 Query 12.2.42 spinners with 1-4 and 8-10; prob product is even
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RESPONSE --> 3/4=0.75
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19:57:05 ** The first number can be 1, 2, 3 or 4. The second can be 8, 9 or 10. There are therefore 4 * 3 = 12 possible outcomes. The only way to get an odd outcome is for the two numbers to both be odd. There are only 2 ways that can happen (1 * 9 and 3 * 9). The other 10 products are all even. So the probability of an even number is 10 / 12 = 5/6 = .833... . Alternatively we can set up the sample space in the form of the table 8 9 10 1 8 9 10 2 16 18 20 3 24 27 30 4 32 36 40 We see directly from this sample space that 10 of the 12 possible outcomes are even. **
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RESPONSE --> ok
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e}dQ~§ assignment #008 cʓ齯̀q}li Liberal Arts Mathematics II 10-11-2006
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20:08:47 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 --> They could be dependent because if the first member chosen was republican, more than likely, the second will also be republican.
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20:08:57 ** 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 --> ok
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20:23:00 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 --> 14/32=0.438 money 13/34=0.382 creativity
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20:23:32 ** 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 --> ok, I figured the wrong things
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20:30:23 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 --> There are four sets, club, diamond, spades, and hearts. If a diamond is dealt first, and club second, this probability is 1/4=0.25
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20:31:06 ** 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 --> ok
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20:32:20 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 --> Well, it could either be a heart or a diamond, so it is 1/2, 0.50
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20:32:29 ** 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 --> ok
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20:38:34 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 --> 0.3 you asked only for thee probability that all 3 sales would be >$100. That is 1/3=0.3
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20:38:49 ** 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 --> ok
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20:40:11 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 --> 0.015= 0.5^5
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20:40:25 ** 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 --> ok, I did it backward
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20:42:29 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 --> It is either 0.04 by finding the difference or 1
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20:42:44 ** 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 --> ok
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20:45:17 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 --> 0.04=0.5*0.8*0.3
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20:46:11 ** 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 --> ok, I see how you got that. I needed to use the information given that 0.8 was the figure given for rain X 2 days
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20:47:19 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 --> 0.8*0.3=0.24
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20:48:00 ** 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 --> Ok, Why would you not use the day that rain had occurred but not the day before?
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20:48:09 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 --> ok
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20:48:16 ** 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 --> ok
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20:48:44 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> I didn't do as well with this section but I think with a little more practice, I should be fine.
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QP~GݴS{° assignment #009 cʓ齯̀q}li Liberal Arts Mathematics II 10-11-2006
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20:52:00 Query 12.4.3 P(2 H on 3 flips)
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RESPONSE --> 2/3*2/3*2/3=8/27=0.30
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20:52:31 ** On three flips you can get HHH, HHT, HTH, HTT, THH, THT, TTH, TTT. Of there 8 possibilities, only 3 of them have two Heads. Thus the probability is 3 / 8. You can get this result without listing. There are 2 possibilities for each flip which gives you 2*2*2 = 2^3 = 8 possible outcomes. To get 2 `heads' you must get `heads' in exactly 2 of the 3 positions. There are C(3, 2) = 3 possible choices of the 3 positions so the probability is C(3,2) / 2^3 = 3/8. If you have n flips, there are C(n,r) ways to get r Heads. This number appears in the n+1 row, as the r+1 entry, of Pascal's triangle. **
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RESPONSE --> ok
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20:53:52 What is the significance of .5^2 * .5 for this question?
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RESPONSE --> I'm not sure how this goes into the previous question
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20:54:54 ** .5^2 is the probability of getting Heads twice in a row. .5 is the probability of a Tails. .5^2 * .5 is therefore the probability of getting HHT. Since the probabilities are independent, you have the same probability of getting two Heads and a Tail in some different order. Since there are C(3,2) possible orders for 2 Heads on 3 coins, the probability of getting 2 Heads and one Tail is C(3,2) * .5^2 * .5 = 3 * .125 = .375, the same as the 3/8 we obtained by listing. **
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RESPONSE --> ok
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