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Math Help - help/confirmation on multi-stage question

  1. #1
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    help/confirmation on multi-stage question

    consider a random experiment done in 2 stages
    first 4 dice are rolled, then a coin is flipped as many times as the number 6 appears

    a)find the probability of getting fewer than 2 heads
    b)knowing the experiment resulted in fewer than 2 heads, what is the conditional probability that exactly 3 sixes were obtained in the first stage

    here is what i have done

    a)
    probability of getting 1 head = (1/12)(11/12)(11/12)(11/12)
    probability of getting 0 head = (11/12)^4
    probability of getting fewer than 2 heads = 0.7703

    b)
    for this, i calculated the probability of getting a certain number of sixes (1, 2, 3 and 4) and then i multiplied those with the probability of getting fewer than 2 heads (respectively)

    i finished with this

    (5/1296)(1/216)/(0.01663) = 0.001074

    I divided by the probability of each set of sixes that would give fewer than 2 heads, that is what the 4 significant digit number is.

    I am somewhat confident in my answer to (a), but and Im not too sure about (b). Help, confirmation?
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  2. #2
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    Quote Originally Posted by chrisc View Post
    consider a random experiment done in 2 stages
    first 4 dice are rolled, then a coin is flipped as many times as the number 6 appears

    a)find the probability of getting fewer than 2 heads
    b)knowing the experiment resulted in fewer than 2 heads, what is the conditional probability that exactly 3 sixes were obtained in the first stage

    here is what i have done

    a)
    probability of getting 1 head = (1/12)(11/12)(11/12)(11/12)
    probability of getting 0 head = (11/12)^4
    probability of getting fewer than 2 heads = 0.7703

    b)
    for this, i calculated the probability of getting a certain number of sixes (1, 2, 3 and 4) and then i multiplied those with the probability of getting fewer than 2 heads (respectively)

    i finished with this

    (5/1296)(1/216)/(0.01663) = 0.001074

    I divided by the probability of each set of sixes that would give fewer than 2 heads, that is what the 4 significant digit number is.

    I am somewhat confident in my answer to (a), but and Im not too sure about (b). Help, confirmation?
    Let X be the random variable number of sixes.

    X ~ Binomial(n = 4, p = 1/6).


    Let Y be the random variable number of heads.

    Y ~ Binomial(n = x, p = 1/2) where x is the possible values of X, that is, 0, 1, 2, 3, 4.


    (a) \Pr(Y < 2) = \Pr(Y = 1) + \Pr(Y = 0).


    \Pr(Y = 1) = \Pr(Y = 1 | X = 1) \cdot \Pr(X = 1) + \Pr(Y = 1 | X = 2) \cdot \Pr(X = 2) + \Pr(Y = 1 | X = 3) \cdot \Pr(X = 3)  + \Pr(Y = 1 | X = 4) \cdot \Pr(X = 1) = \, ....


    \Pr(Y = 0) = \Pr(Y = 0 | X = 0) \cdot \Pr(X = 0) + \Pr(Y = 0 | X = 1) \cdot \Pr(X = 1) + \Pr(Y = 0 | X = 2) \cdot \Pr(X = 2)  + \Pr(Y = 0 | X = 3) \cdot \Pr(X = 3) + \Pr(Y = 0 | X = 4) \cdot \Pr(X = 1) = \, ....


    Therefore \Pr(Y < 2) = \Pr(Y = 1) + \Pr(Y = 0) = \, ....


    The calculations are left for you to do.


    ----------------------------------------------------------------------------------------------------------------------


    (b) \Pr( X = 3 | Y < 2) = \frac{\Pr(Y = 1 | X = 3) \cdot \Pr(X = 3) + \Pr(Y = 0 | X = 3) \cdot \Pr(X = 3)}{\Pr(Y < 2)}.

    Substitute the required probability values from the calculations in part (a).
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  3. #3
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    let me know if i am screwing up the conditional probability because i think i may be

    a)

    y=1:

    (0.06419)(0.09645 + 0.01929 + 0.003858 + 0.09645)
    =0.01387

    y=0:

    (0.7061)(0.4823 + 0.09645 + 0.01929 + 0.003858 + 0.09645)
    =0.4931

    Pr(Y<2) = 0.5070

    --------

    b)

    (0.003858)(0.06419 + 0.7061)/(0.5070) = 0.005862
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  4. #4
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    Quote Originally Posted by chrisc View Post
    let me know if i am screwing up the conditional probability because i think i may be

    a)

    y=1:

    (0.06419)(0.09645 + 0.01929 + 0.003858 + 0.09645)
    =0.01387

    y=0:

    (0.7061)(0.4823 + 0.09645 + 0.01929 + 0.003858 + 0.09645)
    =0.4931

    Pr(Y<2) = 0.5070

    --------

    b)

    (0.003858)(0.06419 + 0.7061)/(0.5070) = 0.005862
    Please explain where each of these numbers have come from because it doesn't look to me like you've calculated the probabilities that are indicated in my first reply.
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  5. #5
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    Pr(y =1) = (1/12)(11/12)^3 = 0.06419
    Pr(y=0) = (11/12)^4 = 0.7061

    Pr(x=0) = (5/6)^4 = 0.4823
    Pr(x=1) = (1/6)(5/6)^3 = 0.09645
    Pr(x=2) = (1/6)^2 x (5/6)^2 = 0.01929
    Pr(x=3) = (1/6)^3 x (5/6) = 0.003858
    Pr(x=4) = (1/6)^4 = 0.0007716

    here is a revision. let me know if this is any better.

    y1 = (1/12)(0.09645)+(11/144)(0.01929)+(121/1728)(0.003858)+(1331/20736)(0.09645) = 0.01597


    y2 = (1)(0.4823)+(11/12)(0.09645)+(121/144)(0.01929)+(1331/1728)(0.003858)+(0.7061)(0.06419) = 0.6352

    y1 + y2 = 0.6512

    Pr(x=3 | y<2) = 0.004979
    Last edited by chrisc; October 16th 2008 at 06:10 AM.
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  6. #6
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    or maybe its like this?

    y1 = (1/2)(x=1) + (2/4)(x=2) + (3/8)(x=3) + (4/16)(x=4)

    the probability of rolling 1 head with 1 six is 1/2
    the probability of rolling 1 head with 2 sixes is 2/4 = 1/2
    the probability of rolling 1 head with 3 sixes is 3/8
    the probability of rolling 1 head with 4 sixes is 4/16 = 1/4

    is that right? well going this way, i ended up with

    Pr(y<2) = 0.5954

    Pr(x=3 | y<2) = 0.003560
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