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April 14th, 2011, 05:08 PM  #1 
Newbie Joined: Apr 2011 Posts: 9 Thanks: 0  Probability... can't seem to get this answer correct.
Hello everyone. This is my first post here and I hope that someone can help me out... I just took a really hard probability test and we have been allowed corrections on the test since everyone did so poorly. I actually did better then most people (78%) and can get a 95% if I do all corrections correctly. I actually feel I did all of them correctly, except for these 2 problems that keep bugging me because I really can't see what I am doing wrong here, I even emailed my instructor and she insisted that I was doing something wrong... So anyway, its two problems, both with a) and b) parts to each problem. Part a) for each problem is WITH REPLACEMENT, meaning that if a number is drawn from a hat, it is placed back into the hat for when the second number is drawn. Part b) for both problems is WITHOUT REPLACEMENT, meaning that if a number is drawn from a hat, it is not placed back into the hat for the second drawing. Anyway, here is the problem: 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? I actually think I did both a) and b) parts correctly for this, however I will write down my corrections and thought process and hopefully someone can tell me if its right or not. Anyways, 'atleast' means that everything must be factored out, and there are 4 possible outcomes in this scenario if we randomly choose 2 numbers: Either Prime&Prime or Prime&Not Prime or Not Prime&Prime or Not Prime&Not Prime. Since the questions says atleast one prime number, the possibilities of NotPrime&NotPrime are excluded from this problem. There are 8 prime numbers and 4 non prime numbers in the set. Therefore if I factor out the problem in the order (P&P or P&NP or NP&P) it should be: (4/12x4/12)+(4/12x8/12)+(8/12x4/12) '&' signifies to multiply, 'or' signifies to add where necessary in equation. Notice how the denominator stays at 12 throughout since part a) is with replacement therefore there will always be 12 numbers to choose from for both first and second drawings. The answer should come out to 80/144. Reduction is not necessary in this test. Part b) is a little different. It is without replacement therefore should look like: (4/12x3/11)+(4/12x8/11)+(8/12x4/11) The second denominator is always 11 since one number has already been drawn and is not put back into the hat. The answer I got was 76/132. Reduction not necessary. Did you get the same answer? I am confident this is right but would like a second opinion! Alright, this one was pretty easy imo. Here's the tough a)/b) question I can't seem to get right: Find the probability that the sum of the two numbers is 30. So this requires more thinking. Starting with part a) first, here are all the possible outcomes where the sum of both numerators sum 30: 2&28, 28&2, 7&23, 23&7, 11&19, 19&11, 13&17, 17&13, 14&16, 16&14, 15&15, 15&15. This is what I wrote down. The only possible error on my part I can conclude is that I am doing 15&15 twice when I should only list them once, however since this is with replacement it seems correct to list 15&15 twice since this is with replacement. Should I only be listing them once instead of twice?? Anyways, assuming you list 15&15 twice, here is the answer I got: (1/12x1/12)+(1/12x1/12)+(1/12x1/12)+(1/12x1/12)+(1/12x1/12)+(1/12x1/12)+(1/12x1/12)+(1/12x1/12)+(1/12x1/12)+(1/12x1/12)+(1/12x1/12)+(1/12x1/12) Answer comes out to 12/144. Reduction not necessary. If 15&15 were only to be counted once the answer would have come out to 11/144. Please help with this! For part b) the sample space is the same but 15&15 are not possible at all since this is without replacement so 15 would only be able to be pulled once. Therefore only 10 possible outcomes 2&28, 28&2, 7&23, 23&7, 11&19, 19&11, 13&17, 17&13, 14&16, 16&14, which would look like: (1/12x1/11)+(1/12x1/11)+(1/12x1/11)+(1/12x1/11)+(1/12x1/11)+(1/12x1/11)+(1/12x1/11)+(1/12x1/11)+(1/12x1/11)+(1/12x1/11) Answer would be 10/132. Reduction not necessary. These are the corrections I have made but I am not confident at all about them as opposed to all the other corrections I made on the other problems I got wrong. Any help would be gladly appreciated! 
April 14th, 2011, 05:36 PM  #2  
Senior Member Joined: Apr 2007 Posts: 2,140 Thanks: 0  Quote:
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. a) with replacement, we have: Thus: This agrees with the result you gave. b) without replacement, we have: Thus: 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: 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: 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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