Results 1 to 3 of 3

Math Help - Old math tests... Argh!

  1. #1
    Junior Member
    Joined
    Dec 2006
    Posts
    40

    Old math tests... Argh!

    So its time! The Final Exam is but two days away!
    I am studying old math tests... but, alas, I did not correct my errors!

    There are 4 problems that i did and I answered incorrectly. I will give my work... Ive looked over them a couple times and (since im rusty from going over them a half year ago) Im not 100% sure what the real thing is.

    1) A basketball team of 10 players - 4guards, 3forwards, 3centers - is electing two captains. Find the probability that a guard And a center will be elected

    so.... 4Guards
    3Forwards
    3Centers
    with 10 in total

    i did {(4C1) X (6C0) / 10C1 } X {(3C1) X (7C0) / 10C1}
    (so it was the Probability for Guards times the probability for Centers )

    and it = .12
    where did i go wrong?


    2) There are 18 juniors and 10 seniors in a class. If 6 of the seniors are males and 12 of the juniors are females, find the probability of selecting a senior or a male

    P(senior) + P(male) - P(Senior and Male)

    so we have 18 juniors with 6M and 12F
    and 10 seniors with 6males and 4 females

    (10/18) + (12/28) - (16/10) = .384126



    3) Find the probability that three cards drawn from a dec without replacement contain at least two hearts

    a deck of card has 52 cards and 13 hearts

    I had 13C2 X 39C1 / 52C3

    Im guessing though, that this one, it was supposed to be
    3C2 X (13/52)^2 X (39/52)^1 since it is WITHOUT replacement, right?


    4) How many ways can a jury of 6men and 6women be selected from 12men and 12 women

    I had 12C6 + 12C6 = 1848
    but im guessing it should have been 12C6 X 12C6 correct?


    Thank you very very much for your time!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,864
    Thanks
    744
    Hello, 3deltat!

    1) A basketball team of 10 players - 4 guards, 3 forwards, 3 centers -
    is electing two captains.
    Find the probability that a guard and a center will be elected.
    Choosing 2 captains from 10 players, there are: / _{10}C_2 \,=\,45 possible choices.

    There are 4 choices for a guard and 3 choices for a center: . 4 \times3 \:=\:12 choices.

    Therefore: . P(\text{guard and center}) \:=\:\frac{12}{45} \:=\:\frac{4}{15}



    2) There are 18 juniors and 10 seniors in a class.
    If 6 of the seniors are males and 12 of the juniors are females,
    find the probability of selecting a senior or a male.
    \begin{array}{ccccccc} & & M & & F & & Total \\ \hline<br />
Juniors & | & 6 &| & 12 & | & 18 \\ \hline<br />
Seniors & | & 6 & | & 4 & | & 10 \\ \hline<br />
Total & | & 12 & | & 16 & | & 28 \\ \hline\end{array}

    Your formula was correct:
    . . P(\text{senior or male}) \:=\:P(\text{senior}) + P(\text{male}) - P(\text{senior and male})

    But the values are: . P(\text{senior or male}) \;=\;\frac{10}{28} + \frac{12}{28} - \frac{6}{28} \;=\;\frac{16}{28} \;=\;\frac{4}{7}



    3) Find the probability that three cards drawn from a deck
    without replacement contain at least two hearts
    There are: . _{52}C_3 = 22,\!100 possible 3-card hands.

    To get exactly three hearts, there are: . _{13}C_3 = 286 ways.

    To get exactly two hearts, there are: . \left(_{13}C_2\right)\left(_{39}C_1\right) = 3042 ways.

    Hence, there are: . 286 + 3042 \:=\:3328 ways to get at least two hearts.


    Therefore: . P(\text{at least 2 hearts}) \;=\;\frac{3328}{22,\!100} \;=\;\frac{64}{425}



    4) How many ways can a jury of 6 men and 6 women be selected from 12 men and 12 women?

    I had: . _{12}C_6 + _{12}C_6 \:= \:1848
    but I'm guessing it should have been: . _{12}C_6 \times\, _{12}C_6, correct? .Right!
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Dec 2006
    Posts
    40
    ah alright. I was looking at your answer for the Without replacement question and was compairing it to a different problem and I think i understand now. What i did the first time, on the test, I made it so that the problem would be trying to find the probability of finding 2 hearts, not "at least" two hearts. Thanks for clearing that up!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Just some tests, trying to use Math Type
    Posted in the LaTeX Help Forum
    Replies: 10
    Last Post: October 11th 2011, 12:43 AM
  2. flux integral.....argh!!!
    Posted in the Calculus Forum
    Replies: 5
    Last Post: May 18th 2010, 04:51 AM
  3. Lagrange Argh...
    Posted in the Calculus Forum
    Replies: 5
    Last Post: October 17th 2009, 08:28 PM
  4. Argh!! Really stuck!! Please help
    Posted in the Differential Equations Forum
    Replies: 5
    Last Post: April 10th 2009, 05:27 AM
  5. Argh
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: October 6th 2005, 09:18 AM

Search Tags


/mathhelpforum @mathhelpforum