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Math Help - simple chance problem

  1. #1
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    simple chance problem

    I need help with figuring out what method to use to figure out this problem..
    Here is the problem.

    A fair coin was flipped 10 times; compute the chance that the number of heads is less than 4.

    Please help me solve this! Thank you.
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  2. #2
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    Hello, emptyglasses!

    You need the Binomial Probability formula . . . several times.


    A fair coin was flipped 10 times.
    Compute the chance that the number of heads is less than 4.

    "Less than 4 heads" means: 3 heads or 2 heads or 1 head or no heads.
    We must find these separate probabilities . . . and add them.

    P(3H)\:=\:\binom{10}{3}\left(\frac{1}{2}\right)^3\  left(\frac{1}{2}\right)^7 \:=\:\frac{120}{1024}

    P(2H) \:=\:\binom{10}{2}\left(\frac{1}{2}\right)^2\left(  \frac{1}{2}\right)^8 \:=\;\frac{45}{1024}

    P(1H) \:=\:\binom{10}{1}\left(\frac{1}{2}\right)^1\left(  \frac{1}{2}\right)^9 \:=\:\frac{10}{1024}

    P(0H)\:=\:\binom{10}{0}\left(\frac{1}{2}\right)^0\  left(\frac{1}{2}\right)^{10}\:=\:\frac{1}{1024}


    Therefore: . P(H < 4) \;=\;\frac{120}{1024} + \frac{45}{1024} + \frac{10}{1024} + \frac{1}{1024} \;=\;\frac{176}{1024} \;=\;\boxed{\frac{11}{64}}

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  3. #3
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    Is there a way of figuring this out without doing all of that computing?
    Another question asked is A fair coin was flipped 100 times, compute the chance that the number of heads is less than 40. I can't imagine doing an equation for every single one through 40!
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  4. #4
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    Quote Originally Posted by emptyglasses View Post
    Is there a way of figuring this out without doing all of that computing?
    Another question asked is A fair coin was flipped 100 times, compute the chance that the number of heads is less than 40. I can't imagine doing an equation for every single one through 40!
    I devised my own approach to this problem.
    Such that can be programmed into a computer.
    I presume there is a way to do it via normal distribution but I never learned probability.

    My approah is to find,
    \int_0^{40} \frac{100!}{40!\cdot \Gamma (41-x)}.5^{100} dx
    Where, s>0
    \Gamma (s)=\int_0^{\infty} e^{-t}t^{s-1}dt (infamous Gamma function).

    I know it looks ugly but programs can do this easily. Taking integral by Simpon's rule. Thus, all you need to do is tell your computer to find this integral.
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  5. #5
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    Quote Originally Posted by emptyglasses View Post
    Is there a way of figuring this out without doing all of that computing?
    Another question asked is A fair coin was flipped 100 times, compute the chance that the number of heads is less than 40. I can't imagine doing an equation for every single one through 40!

    Hi emptyglasses;

    Could i suggest to use Normal distribution method. Normal distribution can be used as an approximation to the binomial distribution (as suggested by soroban) when:
    - Number of trail is large (greater than 30)
    - probability of success at any trail is not too small or large (the closer to 0.5 the better)

    Here you go:

    First you will need to find the expected mean: number of trail x probability of success at any trail = 100 x 0.5 = 50

    Next, find the standard deviation = squaroot 100 x 0.5 x 0.5 = 5

    Then you can find the probability by; P(z < (39.5 - 50)/5) = P(z < -2.1) =0.0179 (find from Normal disribution tables)
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  6. #6
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    Quote Originally Posted by ThePerfectHacker View Post
    I devised my own approach to this problem.
    Such that can be programmed into a computer.
    I presume there is a way to do it via normal distribution but I never learned probability.

    My approah is to find,
    \int_0^{40} \frac{100!}{40!\cdot \Gamma (41-x)}.5^{100} dx.

    Are you sure that is what you use? What about:

    \int_0^{40} \frac{100!}{\Gamma(x+1)\cdot \Gamma (101-x)}(0.5)^{100} dx?

    RonL
    Last edited by CaptainBlack; December 19th 2006 at 04:40 AM.
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  7. #7
    Grand Panjandrum
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    Quote Originally Posted by CaptainBlack View Post
    Are you sure that is what you use? What about:

    \int_0^{40} \frac{100!}{\Gamma(x+1)\cdot \Gamma (101-x)}(0.5)^{100} dx?

    RonL
    It is a nice excecise to compare the normal approximation, the above
    approximation and the exact value. These are 0.0179, 0.0227 and 0.0176
    respectivly.

    Though if we apply a continuity correction to the integral above so we
    integrate from -0.5 to 39.5 the integral changes from 0.0227 to 0.0178
    which is indistinguishable from the normal approximation in this case.


    RonL
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  8. #8
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    Quote Originally Posted by CaptainBlank View Post
    It is a nice excecise to compare the normal approximation, the above
    approximation and the exact value. These are 0.0179, 0.0227 and 0.0176
    respectivly.

    Though if we apply a continuity correction to the integral above so we
    integrate from -0.5 to 39.5 the integral changes from 0.0227 to 0.0178
    which is indistinguishable from the normal approximation in this case.


    RonL
    Thank you for the correction. But my purpose in the post is that it amazingly turns out that we use integration. The reason why the Gamma function appears, as you guessed, is to generalize the factorial. I have a question, is this approach, known? In probability theory.
    Last edited by ThePerfectHacker; December 19th 2006 at 06:29 AM.
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  9. #9
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    Quote Originally Posted by ThePerfectHacker View Post
    Thank you for the correction. But my purpose in the post is that it amazingly turns out that we use integration. The reason why the Gamma function appears, as you guessed, is to generalize the factorial. I have a question, is this approach, known? In probability theory.
    I have a sneaking suspicion that it is virtualy identical to the normal approximation (certainly in this case it appears to be)

    I think the problem is that if you have the computational resources to
    to the integral, you also have the resources to do the exact calculation.

    RonL
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