course Mth 152 I had trouble with two of these assignments and struggled some with a third... as such I figure I had better study this particular area extra well for the test. ŒwìøÙ¸³|ü Éõ™µ…Ƭ^Îyassignment #007
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10:11:33 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 -->
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10:14:15 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 --> The probability is 5/6 My reasoning is this... three total odds (1,3,5) four possibles less than 5 (1,2,3,4) remove the like outcomes (1,3...these two repeat) total correct probable outcomes... 1,2,3,4,5 total outcomes possible... 1,2,3,4,5,6 for 5/6
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10:14:22 ** 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 --> I understand
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10:14:28 Query 12.2.15 drawing neither heart nor 7 from full deck
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10:15:41 Query 12.2.15 drawing neither heart nor 7 from full deck
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RESPONSE --> probability is 9/13 (taken from 36/52( odds in favor is 4 to 9 (from 16 to 36)
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10:15:47 ** 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 --> Exactly
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10:18:58 12.2.24 prob of black flush or two pairs
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RESPONSE --> I had alot of trouble solving this one so I am going to accept your answer and use it to study by.
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10:19:08 ** 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 --> Okay... that makes more sense now.
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10:20:09 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 --> I apologize but I was uncertain about working this one as well... I didn't get enough from the book to figure it out so I want to see your correct response.
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10:20:34 ** 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 --> Okay, I think I understand this a bit more... I will check your cds and see if there is a better explanation on this one.
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10:26:52 Query 12.2.36 n(A)=a, n(S) = s; P(A')=? What is the P(A')?
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RESPONSE --> From what I have learned in the book I think I have the correct understanding of this... for a probability of A' being P(A') I believe the correct breakdown would be something like. n(A')/n(S)= n(A')/n(S)=(s-a)/s this is taken from the fact that A is an event within the sample space of S and we are letting n(A)= a and n(S)=s
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10:27:01 ** 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 --> I think I have it correctly.
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10:30:42 Query 12.2.42 spinners with 1-4 and 8-10; prob product is even
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RESPONSE --> As there are four possible outcomes in the first spinner (aka 1,2,3,4) and three possible outcomes in the second (aka 8,9,10) I would deduce that there are 12 possible outcomes total. As for even numbers I found that most will be even... if multiplying the outcomes together you can get only two odd results, which would be 9 and 27 (taken from 1 * 9 and 3*9). Reducing these from the outcomes, I find a total of 10 possible even outcomes. As such, we find the probability by reducing 10/12 to get 5/6
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10:30:47 ** 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 --> Exactly.
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