Results 1 to 6 of 6

Math Help - Probablitity Help Please.....

  1. #1
    Newbie
    Joined
    May 2009
    Posts
    2

    Probablitity Help Please.....

    My son Is having trouble working out the following:-

    A box contains 2 Strawberry Yougurts, 4 Vanilla yogurts and 6 Cherry Yogurts.

    Three yogurts are selected randomly from the box.

    Calculate the probability that at least one of the selected yogurts is a cherry yogurt.

    Any help greatly appreciated.

    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    Apr 2009
    Posts
    26
    you need to calculate the probability that at least one yogurt is cherry, so that means you need to calculate the probability that one yogurt is cherry, two yogurts are cherry, and three yogurts are cherry and add them together.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,616
    Thanks
    1579
    Awards
    1
    Quote Originally Posted by nightowl007 View Post
    A box contains 2 Strawberry Yougurts, 4 Vanilla yogurts and 6 Cherry Yogurts.
    Three yogurts are selected randomly from the box.
    Calculate the probability that at least one of the selected yogurts is a cherry yogurt.
    Find the the probability of no cherry yogurt: \frac{\binom{6}{3}}{\binom{12}{3}}.

    Substract that number from 1. The oppsite of none is at least one.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    May 2009
    Posts
    2

    Thanks Plato

    Thanks Plato and o&apartyrock for such a quick answers........

    If you can could you simplify answers or explain how you derived your answers?

    Is there a formula to calculate answers for questions like this?

    Thank you very much.

    Nightowl007

    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,616
    Thanks
    1579
    Awards
    1
    There are six non-cherry yougurts, choose three.
    \binom{6}{3}=\frac{6!}{(3!)(3!)}=\frac{(6)(5)(4)}{  (3)(2)(1)}.
    [If your son does not understand this notation, he has no business being asked to do this problem.]
    There are a total of \binom{12}{3} ways to choose three from twelve.
    Last edited by Plato; May 9th 2009 at 01:05 PM.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,711
    Thanks
    630
    Hello, nightowl007!

    I must assume that your son is familiar with Combinations
    . . which has a variety of notations.

    There are 12 objects and we choose 3 of them.
    The number of possible choices can be written: . _{12}C_3\quad\text{ or }{12\choose3}\quad \text{or }C(12,3)

    They are all equal to: . \frac{12!}{3!\,9!} \:=\:220


    A box contains 2 Strawberry yogurts, 4 Vanilla yogurts and 6 Cherry yogurts.

    Three yogurts are selected randomly from the box.

    Calculate the probability that at least one of the selected yogurts is a Cherry yogurt.

    Plato's approach is usually appropriate with an "at least" or "at most" problem.

    There are 6 Cherry yogurts and 6 Others.
    Find the probability that we get no Cherry yogurts.
    . . This means we get three Others.

    There are: . _6C_3 \:=\:20 ways to get 3 Others.

    And there are: . _{12}C_3 \:=\:220 possible outcomes.

    Hence: . P(\text{no Cherry}) \:=\:\frac{20}{220} \:=\:\frac{1}{11}

    Therefore: . P(\text{at least 1 Cherry}) \;=\;1 - \frac{1}{11} \;=\;\boxed{\frac{10}{11}}


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    o&apartyrock's method usually takes longer, but it's correct.
    . . And it is a great way to check our answer.


    1 Cherry (and 2 Others): . (_6C_1)(_6C_2) \:=\:6\cdot15 \:=\:90

    2 Cherries (and 1 Other): . (_6C_2)(_6C_1)  \:=\:16\cdot6 \:=\:90

    3 Cherries (and 0 Others): . (_6C_3)(_6C_0) \:=\: 20\cdot1 \:=\:20


    Hence, there are: .  90 + 90 + 20 \:=\:200 ways to get at least one Cherry.

    Therefore: . P(\text{at least 1 Cherry}) \;=\;\frac{200}{220} \;=\;\frac{10}{11}\quad\hdots\quad \text{ta-}DAA!

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 5
    Last Post: June 21st 2010, 02:31 PM

Search Tags


/mathhelpforum @mathhelpforum