Results 1 to 9 of 9

Math Help - Is this right?

  1. #1
    Junior Member
    Joined
    Mar 2012
    From
    LA
    Posts
    68

    Is this right?

    The way I solved this was (1/5)^6, however that answer came out to be 6.4e-5. However the answer is suppose to be 0.0003200. Can I get a little help on what I did wrong? Tt
    Attached Thumbnails Attached Thumbnails Is this right?-656.png  
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor MarkFL's Avatar
    Joined
    Dec 2011
    From
    St. Augustine, FL.
    Posts
    1,988
    Thanks
    734

    Re: Is this right?

    You found the probability that all six witnesses picked a particular person, not that they all picked the same person. So, let the first witness pick any of the five, then find the probability that the next five witnesses pick that same person.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Sep 2012
    From
    Planet Earth
    Posts
    194
    Thanks
    49

    Re: Is this right?

    After the first witnesses pick a person, each of the remaining 5 witnesses must pick that person, each with probability \frac{1}{5}. So p = \left (\frac{1}{5}\right )^{5} = 0.00032. There's no restriction on who the first person is, just the remaining five must pick the same one.

    Edit - sorry, I misread the above post. Yeah what Mark said is correct.
    Last edited by SworD; September 12th 2012 at 06:43 PM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor MarkFL's Avatar
    Joined
    Dec 2011
    From
    St. Augustine, FL.
    Posts
    1,988
    Thanks
    734

    Re: Is this right?

    Hmmm, since the OP stated P(X)=\left(\frac{1}{5} \right)^6 it appeared to me he did not take that into account...

    edit: LOL, I should have waited 10 seconds before posting again!
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Sep 2012
    From
    Planet Earth
    Posts
    194
    Thanks
    49

    Re: Is this right?

    Yea, I'm developing a habbit of reading posts too quick. I just opened the link, saw the selected choice, and "is this right". Sorry for that xD
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Mar 2012
    From
    LA
    Posts
    68

    Re: Is this right?

    Quote Originally Posted by MarkFL2 View Post
    Hmmm, since the OP stated P(X)=\left(\frac{1}{5} \right)^6 it appeared to me he did not take that into account...

    edit: LOL, I should have waited 10 seconds before posting again!
    Ty, even when I do 1/5^5, it gives me 3.2e-4... Is that a calculator issue(Have TI-84 Silver edition)?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Sep 2012
    From
    Planet Earth
    Posts
    194
    Thanks
    49

    Re: Is this right?

    3.2e-4 = 3.2 \cdot 10^{-4} = 0.00032
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Super Member
    Joined
    Sep 2012
    From
    Washington DC USA
    Posts
    525
    Thanks
    146

    Re: Is this right?

    Another way to see this is that the probability of all 6 witnesses picking a particular man is (1/5)^6. But there are \left( \begin{smallmatrix} 5\\ 1 \end{smallmatrix} \right) = 5 ways to choose a particular man from all 5 men. Therefore the probability of all 6 witness choosing the same man is (5)(1/5)^6 = (1/5)^5 = 0.00032.
    Last edited by johnsomeone; September 13th 2012 at 07:20 PM.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Super Member
    Joined
    Sep 2012
    From
    Washington DC USA
    Posts
    525
    Thanks
    146

    Re: Is this right?

    Yet another completely equivalent way to see the answer is to think of the space of possibilities, which is a set of all the ordered choices of the 6 witnesses:

    Let X = set of possibilities = A \times A \times A \times A \times A \times A = A^6, where A = \{ man_1, man_2, man_3, man_4, man_5 \}.

    Note |X| = |A|^6 = 5^6.

    So, for instance, the element (man_3, man_5, man_2, man_2, man_3, man_4) \in A would correspond to witness #1 choosing man #3,

    witness #2 choosing man #5, witness #3 choosing man #2, and so forth. It's one possibility of the choices of the 6 witnesses.

    Let S be the set of elements of X which represent "all witnesses chose the same man". Then

    S = \{ (man_1, man_1, man_1, man_1, man_1, man_1), (man_2, man_2, man_2, man_2, man_2, man_2), ... (man_5, man_5, man_5, man_5, man_5, man_5) \}.

    Note |S| = 5.

    By the definition of probability, the probability that "all chose the same man" = |S| / |X| = (5) / (5^6) = 1/5^5 = 0.00032.
    Last edited by johnsomeone; September 14th 2012 at 09:20 AM.
    Follow Math Help Forum on Facebook and Google+


/mathhelpforum @mathhelpforum