Results 1 to 5 of 5

Math Help - Organization and multiplication rule

  1. #1
    Newbie
    Joined
    Oct 2012
    From
    Nodak
    Posts
    2

    Organization and multiplication rule

    Hello,

    Ive exhausted all my options on these two problems! Ive been getting everything, just cant figure these two out.

    First Question: (got the first part of it already)

    You are forming a team of five members out of nine people: 4 femals and 5 males.

    b) You need 2 girls and 3 males on your team. In how many ways can you form such a team?

    I might be way off but im thinking

    5 guys, 3 spots: 5 x 4 x 3 = 60
    4 girls, 2 spots: 4 x 3 = 12

    60 x 12 = 720???

    Second problem:

    This is part C of the problem.

    c) Use the multiplication rule to check if the events "being male" and "being a major nurse" are independent.
    Heres the chart.


    -----NursingMajors- NonNMjrs - Total
    Males ---- 95-- ---- 1015------ 1110
    Females 700-- -----1727 ------ 2427
    Total ----795-- -----2742----- --3537

    Thanks!
    Last edited by nodakgolfer; October 9th 2012 at 03:36 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Sep 2012
    From
    Australia
    Posts
    3,612
    Thanks
    591

    Re: Organization and multiplication rule

    Hey nodakgolfer.

    Two events are independent if P(A and B) = P(A)P(B).

    You have all the information in your table, but I'll calculate a few probabilities to get you started (that aren't exactly in the question).

    P(NonNMjrs) = 2742/3537, P(NonNMjrs AND Females) = 1727/3537, P(NursingMajors|Females) = 700/2427.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Oct 2012
    From
    Nodak
    Posts
    2

    Re: Organization and multiplication rule

    I actually have the probabilities down, I'm just wondering what the multiplication rule thing is about and how to use it to see if "being a male" and "being a major nurse" are independent????
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Sep 2012
    From
    Washington DC USA
    Posts
    525
    Thanks
    146

    Re: Organization and multiplication rule

    For part (b)
    How many ways can you choose 3 guys from 5 guys? You calculated it to be 60. The problem is that the way you calculated it respected the order. (60 is the number of ways you can sit 3 guys in 3 differently colored chairs: 5 choices for the red chair, then 4 for the blue chair, then 3 for the green chair.) But since in your case you don't care about order, then you can see that you counted the same "team" of 3 guys multiple times. For a team of Bob, Fred, and Mark, your 60 counted them 6 times (B-F-M, B-M-F, F-B-M, F-M-B, M-F-B, M-B-F). In fact, each team was counted 6 times in that 60. Thus the actual number of male-parts-of-the-team, was 60/6 = 10.

    When choosing where order doesn't matter, we use "combinations", and a special symbol for this because it's so common (note that this special combination notation is *not* a fraction. There is no center bar/numerator, demoninator/etc.. It's just notation.) It's even read as "choose":

    \left(\begin{matrix} 5 \\ 3 \end{matrix} \right) is read "5 choose 3". It equals 5x4x3/6 = 10.

    Recall that you overcounted each team by 6. That's 3!, for the number of ways you can order 3 people (3 for the 1st times 2 for the 2nd, times one for the 3rd/last). You counted all the orderings of a team choice in your 60, so to get to the count of just the number of teams, no ordering, you needed to divide by the number of times all those ordering appeared for the same team.

    That generalizes: "n choose r" = \left(\begin{matrix} n \\ r \end{matrix}\right) = \frac{(n)(n-1)(n-2)...(n-r+1)}{r!} = \frac{n!}{(n-r)! \ r!}.

    That's the number of ways to choose r things from n things, WITH ordering (= (n)(n-1)(n-2)...(n-r+1) = n for the 1st, n-1 for the 2nd, ... (n-r+1) fo the rth/last) *divided* by the the number of orderings that appeared there for each *set* that you choose. When you have a set of r things, it can be ordered r! ways, which is why there's an r! in the demoninator. It's just like the 6 = 3! that needed to divide that 60 by in order to get the right answer.

    One final thing to note is:

    (n)(n-1)(n-2)...(n-r+1)

    = (n)(n-1)(n-2)...(n-r+1)\frac{(n-r)(n-r-1)(n-r-2)...(2)(1)}{(n-r)(n-r-1)(n-r-2)...(2)(1)}

    = \frac{(n)(n-1)(n-2)...(n-r+1)(n-r)(n-r-1)(n-r-2)...(2)(1)}{(n-r)(n-r-1)(n-r-2)...(2)(1)}

    = \frac{n!}{(n-r)!}.

    Thus \frac{(n)(n-1)(n-2)...(n-r+1)}{r!} = \frac{n!}{(n-r)! \ r!}.

    So "n choose r" is:

    \frac{\text{numbers of ways to pick r things from n, with ordering} }{\text{the number of times the same set appears with a different ordering}}

    =\frac{(n)(n-1)(n-2)...(n-r+1)}{r!}, but is written most compactly as \frac{n!}{(n-r)! \ r!}.

    ------------------------------------------------------------------------------------------------------------------
    The number of ways to choose a set of r things out of n things, without ordering, is

    "n choose r" = \left(\begin{matrix} n \\ r \end{matrix}\right) = \frac{n!}{(n-r)! \ r!}.
    ------------------------------------------------------------------------------------------------------------------

    So the number of ways to choose a team of 2 girls, 3 guys, from 4 girls and 5 guys is "4 choose 2 times 5 choose 3"

    = \left(\begin{matrix} 4 \\ 2 \end{matrix}\right) \left(\begin{matrix} 5 \\ 3 \end{matrix}\right) (These are "choose" symbols, not fractions and parentheses)

    = \left(\frac{4!}{(4-2)! \ 2!}\right) \left(\frac{5!}{(5-3)! \ 3!}\right) (These and below *are* fractions and parentheses as usual)

    = \left(\frac{4!}{2! \ 2!}\right) \left(\frac{5!}{2! \ 3!}\right)

    = \left(\frac{(4)(3)(2!)}{2! \ 2!}\right) \left(\frac{(5)(4)(3)(2!)}{2! \ 3!}\right)

    = \left(\frac{(4)(3)}{2!}\right) \left(\frac{(5)(4)(3)}{3!}\right)

    = \left(\frac{(4)(3)}{2}\right) \left(\frac{(5)(4)(3)}{6}\right)

    = (6)(10)

    = 60.
    Last edited by johnsomeone; October 9th 2012 at 07:31 PM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Joined
    Sep 2012
    From
    Australia
    Posts
    3,612
    Thanks
    591

    Re: Organization and multiplication rule

    The definition of independence is P(A and B) = P(A)P(B): if these two are equal, then we consider the events A and B to be independent.

    This definition of independence is used everywhere especially when you need to prove things about independent random variables (like moment generating functions, covariance, etc) as well as for joint distributions (i.e. distributions with more than one random variable) and the biggest one is that the joint distribution of P(A = a, B = b) if A and B are independent is simply P(A = a)P(B = b).

    This is the kind of thing that independence is used for, and it's used for a lot of other models as well as to simplify complex probability models when you have tonnes of variables and you just assume they are independent (which is done in Bayesian data mining stuff, like with the spam filters).
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Multiplication Rule
    Posted in the Discrete Math Forum
    Replies: 4
    Last Post: February 4th 2009, 05:58 PM
  2. Confused about multiplication rule.
    Posted in the Statistics Forum
    Replies: 3
    Last Post: October 7th 2008, 03:14 PM
  3. Multiplication Rule (using dice)
    Posted in the Statistics Forum
    Replies: 3
    Last Post: February 26th 2008, 03:17 PM
  4. special multiplication rule
    Posted in the Statistics Forum
    Replies: 1
    Last Post: January 5th 2008, 10:01 AM
  5. Multiplication Rule
    Posted in the Advanced Statistics Forum
    Replies: 2
    Last Post: May 5th 2006, 01:38 PM

Search Tags


/mathhelpforum @mathhelpforum