My Math Forum Probability : choosing balls (need help)

 July 1st, 2013, 04:28 AM #1 Senior Member   Joined: Apr 2010 Posts: 128 Thanks: 0 Probability : choosing balls (need help) There are nine wrapping bags , three of which contain 2GB pen drives and the rest 1 GB pen drives. There are three girls and five boys randomly select a bag each. Find probability that a) the girls select more bags which contain 2 GB pen drives than the boys, b) none of the girls has a bag which contains 2 GB pen drives. ans given by answer sheet : $a) \ \frac{2}{7} \ \ \ b) \ \frac{5}{28}$ i can't get any of the answer given. even for question b. question a) i got too many type of working , until i confused. b) i go for the girls , $\frac{6}{9}.\frac{3}{8} + \frac{5}{8}.\frac{2}{7} + \frac{4}{7}.\frac{1}{6} \ \= \ \frac{11}{21}$ i wonder are the answers given valid??
 July 1st, 2013, 12:35 PM #2 Global Moderator   Joined: May 2007 Posts: 6,823 Thanks: 723 Re: Probability : choosing balls (need help) For b) I get (6/9)(5/(4/7) = 5/21
 July 1st, 2013, 02:27 PM #3 Senior Member   Joined: Apr 2010 Posts: 128 Thanks: 0 Re: Probability : choosing balls (need help) for the b) because the question said , there are 9 bags lie one the table and the students go and pick them up. The question did not mention about order of the students. so i am under the conception that if all students and bags are put in a container and stir like making milk shake. there would have 3/8 chance for 3/9 2GB bags to meet each other for first try. and once a person pick up , the 2nd one and so forth. but using this concept pose me one major problem....if first bag is for boys and it is 1GB pen drive. The whole probability for girls literally CHANGE such that its decrement is 1 for every value used initially. become (3/7 . 3/ +...and so forth.
 July 2nd, 2013, 06:39 AM #4 Senior Member   Joined: Jun 2013 From: London, England Posts: 1,316 Thanks: 116 Re: Probability : choosing balls (need help) Now, then, this is a very difficult problem, because you've written it down wrong in two key respects! First, there are 8 bags, not 9! Second, you've given the answers to a) and b) in the wrong order!! With the problem as you have stated it, mathman is correct. With 8 bags, the probably that all 3 girls get a 1 GB pen drive becomes: (5/*(4/7)*(3/6) = 5/28 Which you've given as the answer to b). For part b), there are two options: first, that the girls get all the 2 GB drives; second that they get 2 of them. First, the probability that the girls get all 3 is: (3/*(2/7)*(1/6) = 1/56 The probability that the girls get two of them is 3 times the probability that the first two girls get one and the third gets a 1 GB drive, which is: 3*(3/*(2/7)*(5/6) = 15/56 Therefore, the total probability that the girls get more than the boys is 16/56 = 2/7. I expect extra points for getting the right answer, given you got the original problem so muddled up!
 July 2nd, 2013, 06:53 AM #5 Senior Member   Joined: Jun 2013 From: London, England Posts: 1,316 Thanks: 116 Re: Probability : choosing balls (need help) Just realised that the answers are in the right order. But, the main point is that there should be 8 bags not 9 for the answers given.
 July 2nd, 2013, 02:41 PM #6 Senior Member   Joined: Apr 2010 Posts: 128 Thanks: 0 Re: Probability : choosing balls (need help) The original question was a direct extract from my exercise book. No word edit was included. After all no one in school get the right answer even doing it according to a sample question with methods and answer. I see....typing error huh....THX A LOT , my mind is eased a bit burden. We were thinking in school where has our theory gone wrong or is the ans itself are wrg. (never would have thought is the question is having typo)

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