Results 1 to 4 of 4

Thread: Homework Help Please!

  1. Jun 3rd 2009, 11:20 PM #1
    mathsnerd101
    mathsnerd101 is offline
    Newbie
    Joined
    Jun 2009
    Posts
    2

    Homework Help Please!

    I Just got given homework on things i don't get.

    Here are a few questions i'm stuck on:

    Brock has 48 books in his bookcase and 15 of the books are novels. Calculate the probability that of the first 2 books chosen at random for reading (assuming the same book is not chosen twice) that at least one of them is a novel.




    Probability can be used to determine if a gambling game is fair (ie. A fair game is when the cost is equal to the expected winnings)

    EX(winnings)=P(winning)x$amount won
    Expected winnings = Probability of winning x value of prize

    a) If chloe pays $10 to play a game in which she rolls two dice and the sum of the dice is her scroe, determine if this is a fair game if she wins $50 whenever she rolls a 7
    b) Jack was offered the opportunity to play any number of games. In each case he pays $10 to play and roll a pair of dice.

    GAME 1 - if the sum of the numbers is even jack receives $20
    GAME 2 - if the sum of the numbers is 6 or 8 jack receives $30
    GAME 3 - if the sum of the numbers is 5 or less Jak receives $40

    Which Game/s should Jack choose to play? fully justify your answer by comparing all 4 games.



    Thanks to anyone that helps, also i would rather an explanation on how to do it then just an answer. Although how to do it and an answer would be great!
    Follow Math Help Forum on Facebook and Google+
  2. Jun 4th 2009, 04:59 AM #2
    the kopite
    the kopite is offline
    Junior Member
    Joined
    May 2009
    From
    anfield road
    Posts
    29
    first pick probability P= 15/48
    second pick probability P= 15/47
    add that to get the probability of one or the other being a novel. did you not learn the or 'or', 'and' rules?
    Follow Math Help Forum on Facebook and Google+
  3. Jun 4th 2009, 05:25 AM #3
    Soroban
    Soroban is offline
    Super Member
    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    12,028
    Thanks
    847
    Hello, mathsnerd101!

    Brock has 48 books in his bookcase and 15 of the books are novels. .Calculate the
    probability that of the first 2 books chosen at random at least one of them is a novel.

    There are 48 books: . \begin{Bmatrix}\text{15 novels} \\ \text{33 others} \end{Bmatrix}

    There are: . _{48}C_2 \:=\:1128 possible outcomes.

    The opposite of "at least one novel" is "no novels" (all Others).

    No novels: . _{33}C_2 \:=\:528 ways.

    Hence: . 1128 - 528 \:=\:600 ways to get at least one novel.


    Therefore: . P(\text{at least 1 novel}) \;=\;\frac{600}{1128} \;=\;\frac{25}{47}



    a) If Chloe pays $10 to play a game in which she rolls two dice,
    determine if this is a fair game if she wins $50 whenever she rolls a 7
    There are: . 6^2 \:=\:36 possible outcomes.

    There are 6 ways to get a "7": . P(7) \:=\:\tfrac{6}{36} \:=\:\tfrac{1}{6}
    . . She will win $50 one-sixth of the time: . P(+\$50) \:=\:\tfrac{1}{6}

    She pays $10 for every game: . P(\text{-}\$10) \:=\:\tfrac{36}{36} \:=\:1


    Her expected value is: . E \;=\;(50)\left(\tfrac{1}{6}\right) + (\text{-}10)(1) \:=\:-\frac{5}{3}

    She can expect to lose an average of $1.67 per game.



    b) Jack was offered the opportunity to play any number of games.
    In each case he pays $10 to play and roll a pair of dice.

    GAME 1 - if the sum of the numbers is even, Jack receives $20.

    GAME 2 - if the sum of the numbers is 6 or 8, Jack receives $30.

    GAME 3 - if the sum of the numbers is 5 or less, Jack receives $40.

    Which Game/s should Jack choose to play?
    Fully justify your answer by comparing all 4 games. . Four games?
    He pays $10 for every game: . P(-\$10) \:=\:1



    GAME 1 - Half the outcomes are even: . P(+\$20) \:=\:\tfrac{1}{2}

    Therefore: . E \;=\;(20)\left(\tfrac{1}{2}\right) + (-10)(1) \;=\;0

    The game is fair . . . Jack can expect to break even.



    GAME 2 - There are 5 ways to get a "6" and 5 ways to get an "8"
    . . Hence: . P(\text{6 or 8}) \:=\:P(+\$30) \:=\:\tfrac{10}{36} \:=\:\tfrac{5}{18}

    Therefore: . E \;=\;\left(\tfrac{5}{18}\right)(+30) + (1)(-10) \;=\;-\frac{5}{3}

    Jack can expect to lose an average of $1.67 per game.



    GAME 3 - There are 15 outcomes with a sum less than or equal to 5.
    . . Hence: . P(\leq 5) \:=\:P(+\$40) \:=\:\tfrac{15}{36} \:=\:\tfrac{5}{12}

    Therefore: . E \;=\;(+40)\left(\frac{5}{12}\right) + (1)(-10) \;=\;\frac{20}{3}

    Jack can expect to win an average of $6.67 per game.


    It's a no-brainer . . .
    Follow Math Help Forum on Facebook and Google+
  4. Jun 4th 2009, 01:34 PM #4
    mathsnerd101
    mathsnerd101 is offline
    Newbie
    Joined
    Jun 2009
    Posts
    2
    Quote Originally Posted by Soroban View Post
    Hello, mathsnerd101!


    There are 48 books: . \begin{Bmatrix}\text{15 novels} \\ \text{33 others} \end{Bmatrix}

    There are: . _{48}C_2 \:=\:1128 possible outcomes.

    The opposite of "at least one novel" is "no novels" (all Others).

    No novels: . _{33}C_2 \:=\:528 ways.

    Hence: . 1128 - 528 \:=\:600 ways to get at least one novel.


    Therefore: . P(\text{at least 1 novel}) \;=\;\frac{600}{1128} \;=\;\frac{25}{47}



    There are: . 6^2 \:=\:36 possible outcomes.

    There are 6 ways to get a "7": . P(7) \:=\:\tfrac{6}{36} \:=\:\tfrac{1}{6}
    . . She will win $50 one-sixth of the time: . P(+\$50) \:=\:\tfrac{1}{6}

    She pays $10 for every game: . P(\text{-}\$10) \:=\:\tfrac{36}{36} \:=\:1


    Her expected value is: . E \;=\;(50)\left(\tfrac{1}{6}\right) + (\text{-}10)(1) \:=\:-\frac{5}{3}

    She can expect to lose an average of $1.67 per game.



    He pays $10 for every game: . P(-\$10) \:=\:1



    GAME 1 - Half the outcomes are even: . P(+\$20) \:=\:\tfrac{1}{2}

    Therefore: . E \;=\;(20)\left(\tfrac{1}{2}\right) + (-10)(1) \;=\;0

    The game is fair . . . Jack can expect to break even.



    GAME 2 - There are 5 ways to get a "6" and 5 ways to get an "8"
    . . Hence: . P(\text{6 or 8}) \:=\:P(+\$30) \:=\:\tfrac{10}{36} \:=\:\tfrac{5}{18}

    Therefore: . E \;=\;\left(\tfrac{5}{18}\right)(+30) + (1)(-10) \;=\;-\frac{5}{3}

    Jack can expect to lose an average of $1.67 per game.



    GAME 3 - There are 15 outcomes with a sum less than or equal to 5.
    . . Hence: . P(\leq 5) \:=\:P(+\$40) \:=\:\tfrac{15}{36} \:=\:\tfrac{5}{12}

    Therefore: . E \;=\;(+40)\left(\frac{5}{12}\right) + (1)(-10) \;=\;\frac{20}{3}

    Jack can expect to win an average of $6.67 per game.


    It's a no-brainer . . .
    Thank you so much!
    Follow Math Help Forum on Facebook and Google+
« How many 5-card hands make full house | How to figure out binominal distribution »

Similar Math Help Forum Discussions

  1. homework help
    Posted in the Discrete Math Forum
    Replies: 4
    Last Post: Aug 31st 2009, 04:36 AM
  2. need help with homework
    Posted in the Trigonometry Forum
    Replies: 5
    Last Post: Jul 20th 2009, 08:34 AM
  3. homework help
    Posted in the Algebra Forum
    Replies: 2
    Last Post: Feb 10th 2009, 07:12 PM
  4. Homework Help
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Nov 2nd 2007, 01:09 PM
  5. Homework Help
    Posted in the Math Topics Forum
    Replies: 9
    Last Post: Oct 14th 2007, 03:00 PM

Search Tags


/mathhelpforum @mathhelpforum