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Math Help - Binomial Probability

  1. #1
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    Binomial Probability

    In roulette, a steel ball is rolled into a wheel that contains 18 red, 18 black and 2 green slots. If the ball is rolled 25 times, find the probabilities of the following events.

    a) The ball falls into the green slots two or more times
    b) The ball does not fall into any green slots
    c) The ball falls into black slots 15 or more times
    d) The ball falls into red slots 10 or fewer times


    a) so n = 25 .. p = 2/38 and i calculated for x = 0 and x = 1 and added them up and subtracted it from 100% to get 71%.
    b) 36/38 = 95%

    Am i doing this right?
    Can someone help me with c and d, thank you
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by xfyz View Post
    In roulette, a steel ball is rolled into a wheel that contains 18 red, 18 black and 2 green slots. If the ball is rolled 25 times, find the probabilities of the following events.

    a) The ball falls into the green slots two or more times
    b) The ball does not fall into any green slots
    c) The ball falls into black slots 15 or more times
    d) The ball falls into red slots 10 or fewer times


    a) so n = 25 .. p = 2/38 and i calculated for x = 0 and x = 1 and added them up and subtracted it from 100% to get 71%.
    b) 36/38 = 95%

    Am i doing this right?
    Can someone help me with c and d, thank you
    ok, so we are using the binomial distribution, for which the formula is, i'm sure you're familiar with what the variables mean, so i won't define them, P(X = k) \mbox{ or } P(k) = {n \choose k}p^kq^{n - k}

    so for this set of questions, n = 25 always.

    for (a), you want P(X \ge 2) = 1 - P(0) - P(1), here we have p = 2/38. you got that i think. hope your calculations are correct

    for (b) you're not correct. here we want P(0), p = 2/38 also here.

    for (c) you want P(X \ge 15) = P(15) + P(16) + P(17) + ... + P(25). here p = 18/38

    for (d) you want P(X \le 10) = P(0) + P(1) + P(2) + ... + P(10). here p = 18/38
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  3. #3
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    Quote Originally Posted by Jhevon View Post
    ok, so we are using the binomial distribution, for which the formula is, i'm sure you're familiar with what the variables mean, so i won't define them, P(X = k) \mbox{ or } P(k) = {n \choose k}p^kq^{n - k}

    so for this set of questions, n = 25 always.

    for (a), you want P(X \ge 2) = 1 - P(0) - P(1), here we have p = 2/38. you got that i think. hope your calculations are correct

    for (b) you're not correct. here we want P(0), p = 2/38 also here.

    for (c) you want P(X \ge 15) = P(15) + P(16) + P(17) + ... + P(25). here p = 18/38

    for (d) you want P(X \le 10) = P(0) + P(1) + P(2) + ... + P(10). here p = 18/38
    for c and d, do I have to work out each one? For ex: c, do I have to tediously calculate P(15) + p(16) + ... and so on till p(25) all by hand? Is there a faster way?
    thank you
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  4. #4
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    Quote Originally Posted by xfyz View Post
    for c and d, do I have to work out each one? For ex: c, do I have to tediously calculate P(15) + p(16) + ... and so on till p(25) all by hand? Is there a faster way?
    thank you
    Tedious calculations OR look it up in binomial distribution table
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by xfyz View Post
    for c and d, do I have to work out each one? For ex: c, do I have to tediously calculate P(15) + p(16) + ... and so on till p(25) all by hand? Is there a faster way?
    thank you
    as Colby suggested, i guess you could try looking this up in a table. but there is really no other shorter way to do this. the other way would be 1 - P(0) - P(1) - ... - P(14), but that would actually be more calculations.

    there are approximations to the binomial distribution, but i doubt they would be very accurate with n being so small.

    so, i'm guessing the best way for you is to just grit your teeth and plow through it. if you have a calculator which you can save a formula in, it would make life easier for you
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