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Math Help - Homework Help Please!

  1. #1
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    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!
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  2. #2
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    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?
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  3. #3
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    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 . . .

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  4. #4
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    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!
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