course Mth 152 ??i€????y???assignment #008
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18:47:00 `q001. Note that there are 7 questions in this assignment. Suppose that a card is dealt from a well-shuffled deck, and that you can tell by the reflection in your opponent's reading glasses that the card is a red face card. However you can't tell any more than that. What is the probability that the card is the Jack or the Queen of Diamonds?
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RESPONSE --> 1 / 3 confidence assessment: 3
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18:48:23 In this case your knowledge that the card is a red face card limits the possibilities to six: The Jack of Hearts or Diamonds, the Queen of Hearts or Diamonds, or the King of Hearts or Diamonds. The probability that the card is one of the two specified cards is therefore 2 / 6 = 1/3. Note that without any limits on the possibilities, the probability that the card is the Jack or Queen of Diamonds is only 2 / 52 = 1 / 26. Note also that the probability that a card is a red face card is 6 / 52 = 3/26. If we divide the first probability by the second we get 1/26 / ( 3/26) = 1/26 * 26/3 = 1/3. Thus the probability that a card is the Jack or Queen of Diamonds, given that it is a red face card, is equal to the probability that it is the Jack or Queen of Diamonds (and a face card), divided by the probability that it is a red face card. This statement has the form 'The probability of B, given A, is equal to the probability of A ^ B divided by the probability of A'. This statement is abbreviated to the form P(B | A) = P(A ^ B) / P(A). This is the formula for Conditional Probability. In this problem the outcome was Jack or Queen of Diamonds, and the condition was that we have a red face card.
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RESPONSE --> Ok. self critique assessment: 3
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18:50:16 `q002. Suppose that a face card is the first card dealt from a full deck of well-shuffled cards. What is the probability that the next card dealt (without replacement) will also be a face card?
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RESPONSE --> 11 / 51 confidence assessment: 3
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18:50:40 We know that after the first card is dealt there are 11 face cards left out of the original 12, and 51 cards left in the deck. The probability is therefore obviously 11/51. We can also analyze this situation as a conditional probability. B stands for 'a face card is dealt on the second card' while A stands for 'a face card is dealt on the first card'. So the event A ^ B stands for 'a face card is dealt on the first card and on the second', with probability 12/52 * 11/51. A stands for 'a face card is dealt on the first card', with probability 12 / 52. So P(B | A) stands for 'a face card is deal on the second card given that a face card is dealt on the first'. By the formula we have P(B | A) = P ( A ^ B ) / P(A) = [ 12 / 52 * 11 / 51 ] / [ 12 / 52 ] = 11 / 51, which of course we already knew from direct analysis.
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RESPONSE --> Ok. self critique assessment: 3
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18:52:33 `q003. Given that the first clip of a coin is Heads, what is the probability that a five-flip sequence will result in exactly four Heads?
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RESPONSE --> 4 / 16 = 1 / 4 confidence assessment: 3
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18:53:16 If we were to list the 2^5 = 32 possible outcomes for five flips, we would find that 16 of them have 'heads' on the first flip, and that of these 16 there are 4 outcomes with exactly four 'heads'. The probability therefore looks like 4 / 16 = 1/4, which is correct. To verify this by the formula P( B | A ) = P( A ^ B) / P(A), we let B stand for the desired event of exactly four 'heads' and A for the 'given' event of 'heads' on the first flip. On five flips, P(A) = 16 / 32 = 1/2 (probability of 'heads' on the first flip), which P(B ^ A) = 4 / 32 (four of the 32 possible outcomes have 'heads' on the first flip and exactly four 'heads'). The formula therefore gives us P( B | A ) = P( A ^ B) / P(A) = (4/32) / (2/1) = (4 / 32) * (2 / 1) = 4 / 16 = 1/4.
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RESPONSE --> Ok. self critique assessment: 3
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18:59:31 `q004. Given that the first of two dice comes up even, what is the probability that the total on the two dice will be greater than 9? How does this compare with the unconditional probability that the total of two fair dice will be greater than 9?
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RESPONSE --> 4 / 18 = 2 / 9 confidence assessment: 3
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19:00:19 We can list the sample space of dice possibilities for which the first number is even. The sample space is { (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6) }. We note that there are 18 elements in the sample space. We then find the corresponding totals, which are 3, 4, 5, 6, 7, 8 5, 6, 7, 8, 9, 10 7, 8, 9, 10, 11, 12. Of these 18 totals, 4 are greater than 9. Thus the probability that the total of two dice will be greater than 9, given that the first is even, is 4/18 = 2/9. To verify this by the formula P( B | A ) = P( A ^ B) / P(A), we let B stand for the set of all dice pairs which give a total greater than 9, and A for the set of all dice pairs where the first die shows an even number. We have seen that A = { (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6) }. Listing the elements in B we find that B = { (4, 6), (6, 4), (5, 5), (6, 5), (5, 6), (6, 6) }. There are 6 elements in this set. A ^ B consists of the set of elements common to both A and B, or { (4, 6), (6, 4), (6,5), (6, 6) }. Since there are 4 elements in A ^ B, 18 elements in A, and 36 elements in the sample space for two dice, it follows that P(A) = 18 / 36 = 1/2 and P(A ^ B) = 4 / 36 = 1/9. Therefore the probability we are looking for, P(B | A), is given by P(B | A) = P(A ^ B) / P(A) = (1/9) / (1/2) = (1/9) * (2/1) = 2/9. This is in agreement with the previous result obtained by listing.
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RESPONSE --> Ok. self critique assessment: 3
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19:02:28 `q005. A spinner has numbers 2, 3, 4, 5 and 6. Given that the first number is odd, what is the probability that the sum of the results on two consecutive spins is even?
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RESPONSE --> 2 / 5 confidence assessment: 3
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19:04:58 The set of possibilities for which the first number is odd is { (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6) }. There are therefore 10 possibilities. Of these 4 add up to an even total, so the probability that the total is even, given that the first number is odd, is Probability of B given A = 4/10 = 2/5. To verify this by the formula P( B | A ) = P( A ^ B) / P(A), we let B stand for the set of all pairs that add up to an even number and A for the set of all pairs for which the first number is even. The sample space for two spins has 5 * 5 = 25 elements. Of these, only the four outcomes (3, 3), (3, 5), (5, 3) and (5, 5) for which both spinners land on odd numbers are in the set A ^ B. Thus P(A | B) = 4/25. The set A consists of the 10 pairs listed earlier. So P(A) = 10/25 = 2/5. Thus P(B | A) = P(A ^ B) / P(A) = (4/25) / (2/5) = (4/25) * (5/2) = 2/5 in agreement with our previous result.
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RESPONSE --> Ok. self critique assessment: 3
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19:11:47 `q006. What is the probability that two consecutive cards dealt (without replacement) from a full deck will both be Hearts?
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RESPONSE --> C(13,2) = 78 C(52,2) = 1326 78 / 1326 = 1 / 17 confidence assessment: 3
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19:13:25 When the first card is dealt there are 13 Hearts in a deck of 52. When the second card is dealt there are 12 Hearts left in the remaining 51 cards. The order in which the cards are dealt does not matter, and there are two possible orders for any 2-card 'hand'. The number of ways to get 2 Hearts is therefore C(13,2) = 13 * 12 / 2 = 156 / 2 = 78. The number of possible 2-card 'hands' is C(52, 2) = 52 * 51 / 2 = 1326. The probability of obtaining two Hearts is therefore 78 / 1326, which can be reduced or expressed as a decimal.
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RESPONSE --> Ok. self critique assessment: 3
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19:15:36 `q007. What is the probability that two consecutive cards dealt (without replacement) from a full deck will both be of the same suit?
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RESPONSE --> 12 / 51 = 4 / 17 confidence assessment: 3
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19:16:42 A quick common-sense solution tells us that sense the first card can be anything, then since of the 51 remaining cards, there remain remain 12 cards that match the suit of the first the probability must be 12/51 = 4/17.
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RESPONSE --> Ok. self critique assessment: 3
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?????`??~???assignment #009 009. Binomial Probabilities Liberal Arts Mathematics II 06-24-2007
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19:17:55 `q001. Note that there are 5 questions in this assignment. What is the probability that on two rolls of a fair die, we obtain exactly two 3's?
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RESPONSE --> 1 / 36 confidence assessment: 3
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19:18:12 The probability of obtaining a 3 on a single role is 1/6 (one of the six possible outcomes is a 3). Since the two rolls are independent, it follows that if two dice are rolled the probability of obtaining two 3's is 1/6 * 1/6 = 1/36.
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RESPONSE --> Ok. self critique assessment: 3
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19:24:22 `q002. What is the probability that on three rolls of a fair die, we obtain exactly two 5's?
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RESPONSE --> (1 / 6)^2 * (5/6) = 5 / 216 confidence assessment: 2
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19:25:48 On three rolls of a fair die, the two 5's can occur on the first and second, the first and third or the second and third rolls. That is, of the three available positions the two positions in which the 5's occur can occur in C(3,2) = 3 different ways. Since the probability of a 5 on any roll is 1/6 and the probability of not getting a 5 on a roll is 5/6. Any one of the three ways of getting two 5's and one non-5 is therefore (1/6) * (1/6) * (5/6 ) = 5/216. Since each of the three ways to get the desired outcome occurs with probability 5/216, it follows that Probability of exactly two 3's on three rolls = 3 * 5/216 = 15/216 = 5/72.
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RESPONSE --> Ok. self critique assessment: 3
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19:30:46 `q003. What is the probability that on six rolls of a fair die, we obtain exactly two 5's?
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RESPONSE --> (1/6)^2 * (5/6)^4 * C(5,2) confidence assessment: 3
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19:31:18 In order to get exactly two 5's on six rolls of the fair die, we must get two 5's and four results that are not 5. The probability of getting a 5 on any roll is 1/6, and the probability of getting a result other than 5 is 5/6. Therefore given any two positions out of the six the probability of obtaining 5's in two of the positions and non-5's in the remaining four positions is by the Fundamental Counting Principle Probability of 5's in exactly two of the six positions = (1/6) * (1/6) * (5/6) * (5/6) * (5/6) * (5/6) = (1/6)^2 * ( 5/6)^4. There are C(6,2) ways in which the positions of the two 5's can be selected from the six available positions. Thus we have Probability of exactly two 5's on six flips = C(5,2) * (1/6)^2 * (5/6)^4.
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RESPONSE --> Ok. self critique assessment: 3
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19:35:16 `q004. If we let p stand for the probability of getting a 5 on a roll of a die and q for the probability of not getting a 5 on a roll, then how would we expressed a probability of getting exactly r 5's on n rolls?
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RESPONSE --> C(n,r) * (p^r) * q^(n-r) confidence assessment: 3
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19:35:53 By analogy with the preceding problem, we see that to get r 5's on n rolls we must get 5 the total of r times and non-5 a total of (n-r) times. Since probability of getting a 5 is p, the probability of getting 5 a total of r times is represented by p^r. Since the probability of getting a non-5 is q, then the probability of getting a non-5 a total of (n-r) times is represented by q^(n-r). There are C(n, r) ways to place fives in r of n positions, so the probability of getting 5 fives and n non-fives is C(n, r) * p^r * q^(n-r).
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RESPONSE --> Ok. self critique assessment: 3
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19:39:24 `q005. Explain why, if p is the probability of getting a 5 on a single roll of a die, it follows that the probability q of not getting a 5 is q = 1-p. How would we therefore express the formula C(n, r) * p^r * q^(n-r) only in terms of p?
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RESPONSE --> confidence assessment:
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19:39:40 If we roll a single die, we either get 5 or we don't. The two events are mutually exclusive -- they can both happen on the same roll. They also cover all possibilities. The sum of the probabilities is therefore 1. So we conclude that p + q = 1, and from this it follows immediately that q = 1 - p. Substituting 1 - p for q in the expression C(n, r) * p^r * q^(n-r) we obtain Probability of r fives on n rolls = C(n, r) * p^r * (1-p) ^ (n-5).
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RESPONSE --> self critique assessment:
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??C????\????????? assignment #008 008. `query 8 Liberal Arts Mathematics II 06-24-2007
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21:05:45 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 are dependant because the odds of the second choice depends on which member was chosen first.
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21:08:17 ** 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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21:13:00 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 --> 27 / 66 = 9 / 22
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21:15:42 ** 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.
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21:31:08 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 --> 13 / 51 since 13 clubs remain in the deck of 51
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21:31:23 ** 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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21:40:55 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 --> 13 / 26 = 1 /2
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21:43:19 ** 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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21:51:35 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 = .512
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21:54:02 ** 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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22:27:14 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 --> .95^5 = .774
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22:27:45 ** 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.
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22:31:47 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 = .250
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22:34:53 ** 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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22:37:02 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 * .8^3 = .154
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22:38:40 ** 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.
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22:41:26 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 --> .5 * .8^2 = .32
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22:42:02 ** 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.
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22:44:26 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 isn't included since it isn't odd.
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22:44:53 ** 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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22:45:05 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> None.
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z??????????f???assignment #009 009. `Query 9 Liberal Arts Mathematics II 06-24-2007
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22:52:19 Query 12.4.3 P(2 H on 3 flips)
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RESPONSE --> 3 / 8
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22:54:08 ** 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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22:55:40 What is the significance of .5^2 * .5 for this question?
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RESPONSE --> The probability of getting heads, heads, tails.
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22:57:08 ** .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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23:11:10 Query 12.4.6 P(>= 1 H on 3 flips) Give the requested probability and explain how you obtained your result.
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RESPONSE --> .5^3 = .125
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23:11:30 ** Probability of getting no heads on three flips is P(TTT) = .5 * .5 * .5 = .125, or 1/8, obtained by multiplying the probability of getting a tails for each of 3 independent flips. Subtracting this from 1 gives .875, or 7/8. **
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RESPONSE --> Ok.
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23:18:23 Query 12.4.15 P(3 H on 7 flips) Give the requested probability and explain how you obtained your result.
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RESPONSE --> C(7,3) = 35 .5^3 * .5^4 = 1 / 128 35 * 1 / 128 = 35 / 128
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23:19:31 ** There are C(7,3) = 7 * 6 * 5 / 3! = 35 ways to choose three of the 7 `positions' for Heads on 7 flips. So there are C(7,3) = 7 * 6 * 5 / 3! = 35 ways to get three heads on 7 flips. The probability of any of these ways is (1/2)^3 * (1/2)^4 = 1 / 2^7 = 1 / 128. The probability of 3 Heads on 7 flips is therefore 35 * 1/128 = 35 / 128. **
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RESPONSE --> Ok.
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23:21:29 Query 12.4.21 P(1 success in 3 tries), success = 4 on fair die
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RESPONSE --> C(3,1) * 1/6 * (5/6)^2 = 4 * 1/6 * 25 / 36 = 100 / 216 = 25 / 72
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23:22:03 ** To get 1 success on 3 tries you have to get 1 success and 2 failures. On any flip the probability of success is 1/6 and the probability of failure is 5/6. For any ordered sequence with 1 success and 2 failures the probability is 1/6 * (5/6)^2. Since there are C(3,1) = 3 possible orders in which exactly 1 success can be obtained, the probability is C(3,1) * 1/6 * (5/6)^2 = 4 * 1/6 * 25 / 36 = 100 / 216 = 25 / 72. This matches the binomial probability formula C(n, r) * p^r * q^(n-r), with prob of success p = 1/6, prob of failure q = 1 - 1/6 = 5/6, n = 3 trials and r = 1 success. **
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RESPONSE --> Ok.
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23:23:18 Query 12.4.33 P(exactly 7 correct answers), 3-choice mult choice, 10 quest. What is the desired probability?
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RESPONSE --> C(10,7) * (2/3)^3 * (1/3)^7
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23:24:31 ** The probability of a correct answer from a random choice on any single question is 1/3. For any sequence of 7 correct answers and 3 incorrect the probability is (1/3)^7 * (2/3)^3. There are C(10,7) possible positions for 7 correct answers among 10 questions. So the probability is C(10,7) * (1/3)^7 * (2/3)^3 = 320/19683 = 0.0163 approx. This matches the binomial probability formula C(n, r) * p^r * q^(n-r), with prob of success p = 1/3, prob of failure q = 1 - 1/3 = 2/3, n = 10 trials and r = 7 success. ANOTHER SOLUTION: There are C(10,7) ways to distribute the 7 correct answers among the 10 questions. The probability of any single outcome with 7 successes and 3 failures is the product of (1/3)^7, representing 7 successes, and (2/3)^3, representing 3 failures. The probability of exactly seven correct questions is therefore prob = C(10,7) * (2/3)^3 * (1/3)^7 . **
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RESPONSE --> Ok.
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23:28:53 Query 12.4.39 P(more than 2 side effect on 8 patients), prob of side effect .3 for each
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RESPONSE -->
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23:29:02 ** The probability of 0 side effects is C(8,0) * .7^8. The probability of 1 side effect is C(8,1) * .7^7 * .3^1. The probability of 2 side effects is C(8,2) * .7^6 * .3^2. The sum of these two probabilities is the probability that two or fewer patients will have side effects. We subtract this probability from 1 to get the probability that more than 2 will experience side effects. The result is approximately .448. DER**
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RESPONSE -->
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23:31:47 Query 12.4.48 P(4 th child is 1 st daughter) What is the probability that the fourth child is the first daughter and how did you obtain your result?
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RESPONSE --> .5^3 * .5 = .063
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23:32:23 ** The fourth child will be the first daughter if the sequence is SSSD, S standing for son and D for daughter. The probability of S on any birth is .5, and the probability of G is .5. The probability of SSSD is .5^3 * .5 = .0625 or 1/16. **
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RESPONSE --> Ok.
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23:36:52 Query 12.4.54 10-step rnd walk, 1 dim; P(6 South) What is the probability of ending up 6 blocks South of the starting point and how did you obtain it?
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RESPONSE --> C(10,8) * .5^8 * .5^2 = .044
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23:38:13 ** To end up 6 blocks South requires 8 steps South and 2 steps North. Thus exactly 8 of the 10 steps must be South, and there are C(10,8) ways for this to happen. The probability of any given combination of 8 South and 2 North is (1/2)^8 * (1/2)^2 = 1 / 2^10 = 1 / 1024. The probability of ending up 6 blocks South is therefore prob = C(10,8) * (1/2)^8 * (1/2)^2 = 45 * (1/2)^10 = 45 / 1024, or about .044. **
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RESPONSE --> Ok.
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23:39:29 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> None.
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