My Math Forum Probability... can't seem to get this answer correct.

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April 14th, 2011, 05:36 PM   #2
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Quote:
 Originally Posted by foliocb Each of the numbers {2,5,7,11,13,14,15,16,17,19.23.28} is put into a hat. Two of the numbers are selected a) with replacement, b) without replacement. 1.) What is the probability that atleast one number is not a prime?
Prime = {2, 5, 7, 11, 13, 17, 19, 23} (8 numbers)
Not prime = {14, 15, 16, 28} (4 numbers)
Total of 12 numbers
There are three possibilities here: Two numbers are both prime, neither prime, or one prime other not.
I assume you are smart enough to already know these, you think you could solve it completely?

 April 14th, 2011, 05:44 PM #3 Newbie   Joined: Apr 2011 Posts: 9 Thanks: 0 Re: Probability... can't seem to get this answer correct. Yes, however my instructor always says 'order is important' and since there are only two outcomes(prime or not prime), and order is important, then NP & P exists but also so does the possibility P & NP... so there's actually 4 possibilities, since the question states that atleast one # must be not prime then P&P is excluded. Think I made a typo above and said NP&NP is excluded. Sorry about that... the equation is still correct though. So what do you mean? Work must be shown here...
 April 14th, 2011, 05:49 PM #4 Senior Member   Joined: Apr 2007 Posts: 2,140 Thanks: 0 If you're really not rude and smarter than I am mathematically, then show us your work!
 April 14th, 2011, 05:53 PM #5 Newbie   Joined: Apr 2011 Posts: 9 Thanks: 0 Re: Probability... can't seem to get this answer correct. Not being rude... lol more like confused. (4/12x4/12)+(4/12x8/12)+(8/12x4/12) = 80/144. NP x NP + NP x P + P x NP = That's the layout of the equation... is that what you meant?
 April 14th, 2011, 05:56 PM #6 Senior Member   Joined: Apr 2007 Posts: 2,140 Thanks: 0 As long you're not being rude that's good. I'm confused on the problem too.
 April 14th, 2011, 05:58 PM #7 Newbie   Joined: Apr 2011 Posts: 9 Thanks: 0 Re: Probability... can't seem to get this answer correct. A way for proof: We can also do NPxNP+NPxP+PxNP+PxP = 1... 1 being 144 total. So therefore, (PxP) = (8/12x8/12) = 64/144. 80+64 = 144.
 April 14th, 2011, 06:09 PM #8 Senior Member   Joined: Apr 2007 Posts: 2,140 Thanks: 0 Looks good to me! MarkFL, what do you think?
 April 14th, 2011, 06:43 PM #9 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Probability... can't seem to get this answer correct. Here's how I would do these problems: 1. Each of the numbers {2,5,7,11,13,14,15,16,17,19,23,28} is put into a hat. Two of the numbers are selected a) with replacement, b) without replacement. What is the probability that at least one number is not a prime? We can use the complementation rule. Let P(A) be the probability that both are primes. Then P(not A) will be equivalent to the probability that at least one is not a prime. We have 8 primes, and 4 composites. $P$$\text{not }A$$=1-P(A)$ a) with replacement, we have: $P(A)=$$\frac{8}{12}$$^2=$$\frac{2}{3}$$^2=\frac{4} {9}$ Thus: $P$$\text{not }A$$=1-\frac{4}{9}=\frac{5}{9}$ This agrees with the result you gave. b) without replacement, we have: $P(A)=\frac{8}{12}\cdot\frac{7}{11}=\frac{14}{33}$ Thus: $P$$\text{not }A$$=1-\frac{14}{33}=\frac{19}{33}$ This agrees with the result you gave. 2. Find the probability that the sum of the two numbers is 30. a) with replacement, there are, by the product rule, 12² = 144 outcomes. The outcomes where the sum is 30 are: {2,28}, {7,23}, {11,19}, {13,17}, {14,16}, {15,15}, {16,14}, {17,13}, {19,11}, {23,7}, {28,2} (I counted {15,15} once since it is the same draw) Thus, there are 11 outcomes where the sum is 30, so the probability is: $P(A)=\frac{11}{144}$ b) without replacement Now there are 12·11 = 132 total outcomes, and we cannot draw 15 twice, so we only have 10 favorable outcomes, thus: $P(A)=\frac{10}{132}=\frac{5}{66}$ This agrees with the result you gave.
 April 14th, 2011, 07:47 PM #10 Newbie   Joined: Apr 2011 Posts: 9 Thanks: 0 Re: Probability... can't seem to get this answer correct. Thanks. 11/144 seems like a better answer then my 12/144.

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