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Math Help - Probabilty Confusion

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    Probabilty Confusion

    A red and a black die labelled 1 - 6 are thrown simultaneously, find the probability of:

    A number less than 5 on the red die but a total of 10 altogether.

    I know the answer is 1/36... however i do not understand why there wasnt an inclusion of a restricted sample space. e.g. Pr(Total 10 | <5 on red die) where: Pr (Total 10 And <5 on red) / Pr(>5 on red) = (1/36)/(4/6) = 1/24

    Similarly, A pair of fair dice are rolled. Find the probability that both numbers are odd given that the first die shows a number less than 4.

    How come this one involves an inclusion of a restricted sample space?

    Please explain in simple terms guys, im not too advance in mathematical terminology. Thnks.



    Last edited by andrew2322; April 29th 2008 at 03:03 AM.
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    Hello,

    As an answer to your PM...

    For the first question, you just want P(Total 10 And <5 on red).

    The text doesn't say "the probability to get 10 IF the red dice <5". It's "get 10 with red < 5"

    The only possibility is red dice=4 and black dice=6, so it's 1/36, since you have :

    P(Total 10 And <5 on red)=P(Red dice=4)*P(Black dice=6)
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    Similarly, A pair of fair dice are rolled. Find the probability that both numbers are odd given that the first die shows a number less than 4.
    The possibilities for the first dice are 1 and 3.
    So the probability that the first dice is odd is 2/6.

    The possibilities for the second dice are 1, 3 and 5.
    So the probability that the second dice is odd is 3/6.


    P(both dices are odd)=P(1st dice odd AND 2nd dice odd)=P(1st dice odd)P(2nd dice odd)=...
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    hey

    In the book, the second one was under the conditional probabiltiy section, im so confused ahhhhhh pls HELP also i do not understand ur working out pls help me... IM STUPID
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    Quote Originally Posted by andrew2322 View Post
    In the book, the second one was under the conditional probabiltiy section, im so confused ahhhhhh pls HELP
    I don't really understand why you want to use conditional probability since the events are independent...
    Perhaps there is a way to do it, but then, you asked the wrong person
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    ok ok, hold on just one more thing. i got 6/36 or 1/6 for question 2 ... ok now, when i think about it logically isnt the 36 reduced because the first has to be less than 4 and cant be 5 or 6?
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    Quote Originally Posted by andrew2322 View Post
    ok ok, hold on just one more thing. i got 6/36 or 1/6 for question 2 ... ok now, when i think about it logically isnt the 36 reduced because the first has to be less than 4 and cant be 5 or 6?
    Hm, not really, there is no relation. We're talking about odd numbers, so you can also say that you're working on 3 possibilities only. I'm really sorry, this kind of things is really hard for me to explain in English...

    also i do not understand ur working out pls help me... IM STUPID
    No one is stupid. Tell me what's wrong
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    hey

    hey i really jsut dont understand how come the 36 isnt affected by only using numbers less than 4 on the first die? and with your working, i do not understand after the first equals sign P(both dices are odd)=P(1st dice odd AND 2nd dice odd)=P(1st dice odd)P(2nd dice odd)=...

    and is my answer correct? 1/6

    Also i have one more problem that i need help with
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    Quote Originally Posted by andrew2322 View Post
    hey i really jsut dont understand how come the 36 isnt affected by only using numbers less than 4 on the first die?
    Because you're working on probabilities to get the first dice < 4. The probability will be related to 6 since there are 6 possibilities but only 3 interest you (<4), and only 2 really interest you (odd numbers).


    and with your working, i do not understand after the first equals sign P(both dices are odd)=P(1st dice odd AND 2nd dice odd)=P(1st dice odd)P(2nd dice odd)=...
    numbers on both dices are odd means that each dice has to figure out an odd number. So first dice has to be odd, and the second one too

    and is my answer correct? 1/6
    Yes it is
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    hey

    In a school of 100 students, 85 are atctive participants in hockey or basketball. Basketball is popular with 75 players, but the coach is happy with the squad of 15. Find the probability that a student chosen at random is:

    a.) A hockey player
    b.) A participant in both sports
    c.) Not a basketballer
    d.) Not a participant in either sport

    for this problem i attempted to draw up a venn diagram, however i could not figure out all the values, please help me. Please reply with answers, and working out i have to go to sleep for school tomorrow, thanks
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    Quote Originally Posted by andrew2322 View Post

    [snip]
    Similarly, A pair of fair dice are rolled. Find the probability that both numbers are odd given that the first die shows a number less than 4.

    How come this one involves an inclusion of a restricted sample space?

    Please explain in simple terms guys, im not too advance in mathematical terminology. Thnks.
    Consider the reduced sample space:

    (1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)
    (2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)
    (3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)

    18 possible outcomes, 6 are favourable ......
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    In a school of 100 students, 85 are atctive participants in hockey or basketball. Basketball is popular with 75 players, but the coach is happy with the squad of 15. Find the probability that a student chosen at random is:

    a.) A hockey player
    b.) A participant in both sports
    c.) Not a basketballer
    d.) Not a participant in either sport

    for this problem i attempted to draw up a venn diagram, however i could not figure out all the values, please help me. Please reply with answers, and working out i have to go to sleep for school tomorrow, thanks
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  13. #13
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    Quote Originally Posted by andrew2322 View Post
    In a school of 100 students, 85 are atctive participants in hockey or basketball. Basketball is popular with 75 players, but the coach is happy with the squad of 15. Find the probability that a student chosen at random is:

    a.) A hockey player
    b.) A participant in both sports
    c.) Not a basketballer
    d.) Not a participant in either sport

    for this problem i attempted to draw up a venn diagram, however i could not figure out all the values, please help me. Please reply with answers, and working out i have to go to sleep for school tomorrow, thanks
    "85 are atctive participants in hockey or basketball." => 15 play neither.
    "85 are atctive participants in hockey or basketball. Basketball is popular with 75 players" => 10 play hockey only.
    "the coach is happy with the squad of 15" => 15 play basketball only??
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    Sorry

    Sorry, the hockey coach is happy with the squad of 15, please demonstrate using a venn diagram.
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  15. #15
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    Quote Originally Posted by andrew2322 View Post
    Sorry, the hockey coach is happy with the squad of 15, please demonstrate using a venn diagram.
    Then since 10 play hockey only, 5 play hockey and basket ball.
    5 play hockey and basketball and "Basketball is popular with 75 players" => 70 play basketball only.
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