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Math Help - Probability Help please guys...

  1. #1
    Bar0n janvdl's Avatar
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    Probability Help please guys...

    Due to a math test I could not go to the extra class on Friday to get help on my tutorial, so I thought I'd ask here...

    If X is a geometric random variable with p = 0,5 ; for what value of k is P(X \leq k) \approx 0,99
    My solution:

    \sum_{k=1}^{n} \left(  (1-p)^{k-1} (p)  \right) \approx 0,99

    \sum_{k=1}^{n} \left(  (1-0,5)^{k-1} (0,5)  \right) \approx 0,99

    \sum_{k=1}^{n} \left(  (0,5)^{k-1+1}  \right) \approx 0,99

    \sum_{k=1}^{n} \left(  (0,5)^{k}  \right) \approx 0,99

    Okay now I got n = 7. But to me it is a bit of a elementary approach to just keep changing the value for n and seeing what I get. Is there a method to solve this?


    Three identical fair coins are thrown simultaneously until all three show the same face. What is the probability that they are thrown more than three times?
    My solution:

    Okay now I used negative binomial distribution on this. The thing is we want 3 successes, so r = 3. I was struggling with thinking of a value for k, but logic tells me it must be a multiple of 3?

    P(X > 3) = 1 - P(X \leq 3)

    1 - {3-1 \choose 3-1} (0,5)^3 (1-0,5)^{3-3} - {6-1 \choose 3-1} (0,5)^3 (1-0,5)^{6-3} - {9-1 \choose 3-1} (0,5)^3 (1-0,5)^{9-3}

    Suppose the lifetime of an electronic component follows an exponential distribution with \lambda = 0,1

    a) Find the probability that the lifetime is less than 10
    b) Find the probability that the lifetime is between 5 and 15
    c) Find t such that the probability that that the lifetime is greater than t is 0,1
    My solution:

    a) \sum_{t=1}^{9} e^{(-0,1)(t)}

    b) e^{(-0,1)(15)} - e^{(-0,1)(5)}

    c) 1 - \sum_{t=1}^{n} e^{(-0,1)(t)} = 0,01

    \sum_{t=1}^{n} e^{(-0,1)(t)} = 0,99

    Okay something isn't right with my number c. If i let n = 1 I get the probability is 0,90 but if I let n = 2 it ends up greater than one.

    If U is uniform on [0;1] find the density function of \sqrt{U}
    My solution:

    Let U be a uniform random variable on [0;1], and let V = \sqrt{U}

    F_{V}(v) = P(V \leq v)

    = P(\sqrt{U} \leq v)

    = P(U \leq v^2)

    = f_{U} (v^2)

    = v^2

    Differentiate.

    f_{V} (v) = 2v

    A particle of mass M has a random velocity, V, which is normally distributed with parameters \mu = 0 and \sigma. Find the density function of the kinetic energy.
    Just when I thought i was done with physics forever...

    My solution:

    E = \frac{1}{2} m V^2

    F_{E} (e) = P(E \leq e)

    = P \left( \frac{1}{2}mV^2 \leq e \right)

    = P \left( V \leq \sqrt{\frac{2e}{m}} \right)

    = f_{V} \left( \sqrt{\frac{2e}{m}} \right)

    = \sqrt{\frac{2e}{m}}

    Differentiate. (Assuming the mass is a constant)

    \frac{d}{de} = \frac{1}{2 \sqrt{\frac{2e}{m}}}

    If the radius of a circle is an exponential random variable, find the density function of the area
    What the freak are these people on about??

    My solution:

    A = \pi r ^2

    F_{A} (a) = P(A \leq a)

    = P \left( \pi r ^2 \leq a \right)

    = P \left( r \leq \sqrt{\frac{a}{\pi}} \right)

    = f_{r} \left( \sqrt{\frac{a}{\pi}} \right)

    = \sqrt{\frac{a}{\pi}}

    Differentiate.

    \frac{d}{da} = \frac{1}{2 \sqrt{\frac{a}{\pi}}}

    ======

    Thanks in advance
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  2. #2
    Flow Master
    mr fantastic's Avatar
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    Quote Originally Posted by janvdl View Post
    Due to a math test I could not go to the extra class on Friday to get help on my tutorial, so I thought I'd ask here...



    My solution:

    \sum_{k=1}^{n} \left( (1-p)^{k-1} (p) \right) \approx 0,99

    \sum_{k=1}^{n} \left( (1-0,5)^{k-1} (0,5) \right) \approx 0,99

    \sum_{k=1}^{n} \left( (0,5)^{k-1+1} \right) \approx 0,99

    \sum_{k=1}^{n} \left( (0,5)^{k} \right) \approx 0,99

    Okay now I got n = 7. But to me it is a bit of a elementary approach to just keep changing the value for n and seeing what I get. Is there a method to solve this?




    My solution:

    Okay now I used negative binomial distribution on this. The thing is we want 3 successes, so r = 3. I was struggling with thinking of a value for k, but logic tells me it must be a multiple of 3?

    P(X > 3) = 1 - P(X \leq 3)

    1 - {3-1 \choose 3-1} (0,5)^3 (1-0,5)^{3-3} - {6-1 \choose 3-1} (0,5)^3 (1-0,5)^{6-3} - {9-1 \choose 3-1} (0,5)^3 (1-0,5)^{9-3}



    My solution:

    a) \sum_{t=1}^{9} e^{(-0,1)(t)}

    b) e^{(-0,1)(15)} - e^{(-0,1)(5)}

    c) 1 - \sum_{t=1}^{n} e^{(-0,1)(t)} = 0,01

    \sum_{t=1}^{n} e^{(-0,1)(t)} = 0,99

    Okay something isn't right with my number c. If i let n = 1 I get the probability is 0,90 but if I let n = 2 it ends up greater than one.



    My solution:

    Let U be a uniform random variable on [0;1], and let V = \sqrt{U}

    F_{V}(v) = P(V \leq v)

    = P(\sqrt{U} \leq v)

    = P(U \leq v^2)

    = f_{U} (v^2)

    = v^2

    Differentiate.

    f_{V} (v) = 2v



    Just when I thought i was done with physics forever...

    My solution:

    E = \frac{1}{2} m V^2

    F_{E} (e) = P(E \leq e)

    = P \left( \frac{1}{2}mV^2 \leq e \right)

    = P \left( V \leq \sqrt{\frac{2e}{m}} \right)

    = f_{V} \left( \sqrt{\frac{2e}{m}} \right)

    = \sqrt{\frac{2e}{m}}

    Differentiate. (Assuming the mass is a constant)

    \frac{d}{de} = \frac{1}{2 \sqrt{\frac{2e}{m}}}



    What the freak are these people on about??

    My solution:

    A = \pi r ^2

    F_{A} (a) = P(A \leq a)

    = P \left( \pi r ^2 \leq a \right)

    = P \left( r \leq \sqrt{\frac{a}{\pi}} \right)

    = f_{r} \left( \sqrt{\frac{a}{\pi}} \right)

    = \sqrt{\frac{a}{\pi}}

    Differentiate.

    \frac{d}{da} = \frac{1}{2 \sqrt{\frac{a}{\pi}}}

    ======

    Thanks in advance
    I have no time now. But will answer all these a bit later.
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  3. #3
    Bar0n janvdl's Avatar
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    Quote Originally Posted by mr fantastic View Post
    I have no time now. But will answer all these a bit later.
    Thank you, Mr F
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  4. #4
    Flow Master
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    Here're my thoughts on the first couple:

    Quote Originally Posted by janvdl View Post
    [snip]
    Quote:
    If X is a geometric random variable with p = 0,5 ; for what value of k is

    The cdf for a geometric distribution is {\color{red}\Pr(X \leq k) = 1 - (1 - p)^k}.

    So you're required to solve {\color{red}1 - \left( \frac{1}{2}\right)^k \approx 0.99 } for k. This obviously boils down to solving {\color{red} \left( \frac{1}{2}\right)^k \approx 0.01 .......}

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

    Quote:
    Three identical fair coins are thrown simultaneously until all three show the same face. What is the probability that they are thrown more than three times?

    My solution:

    Okay now I used negative binomial distribution on this.

    The geometric distribution should be used here. The probability of success (getting three faces the same) in a single trial (three coins tossed) is {\color{red} \frac{1}{8} + \frac{1}{8} = \frac{1}{4}}.

    You want {\color{red} \Pr(X > 3) = 1 - \Pr(X \leq 3) = 1 - \left[ 1 - \left( 1 - \frac{1}{8}\right)^3 \right] = \left( \frac{7}{8}\right)^3}.

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

    Quote:
    Suppose the lifetime of an electronic component follows an exponential distribution with

    a) Find the probability that the lifetime is less than 10
    b) Find the probability that the lifetime is between 5 and 15
    c) Find such that the probability that that the lifetime is greater than is 0,1

    The pdf of an exponential distribution is {\color{red} f(x) = \lambda e^{-\lambda x}}.

    a) {\color{red} \Pr(X < 10) = \int_{0}^{10} f(x) \, dx = .....}

    b) {\color{red} \Pr(5 < X < 15) = \int_{5}^{15} f(x) \, dx = .....}

    c) Solve the equation {\color{red} \Pr(X > t) = \int_{t}^{+\infty} f(x) \, dx = 0.1} ~ for t (well, there'll be an equation once you do the integration ).

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

    Quote:
    If is uniform on [0;1] find the density function of

    My solution:

    Let U be a uniform random variable on [0;1], and let V = \sqrt{U}

    F_{V}(v) = P(V \leq v)

    = P(\sqrt{U} \leq v)

    = P(U \leq v^2)

    = f_{U} (v^2)

    = v^2

    Differentiate.

    f_{V} (v) = 2v for {\color{red} 0 \leq v \leq 1} and f(v) = 0 otherwise.

    This is good! But you should add the bit in red too.
    [snip]
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  5. #5
    Flow Master
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    Quote Originally Posted by janvdl View Post
    [snip]
    Quote:
    A particle of mass has a random velocity, , which is normally distributed with parameters and . Find the density function of the kinetic energy.

    [snip]

    Quote:
    If the radius of a circle is an exponential random variable, find the density function of the area
    For these two you're on the wrong track I'm afraid.

    In each case the problem is more or less the following:

    The pdf for X is f(x), find the pdf of aX^2.

    Have a look at this thread: http://www.mathhelpforum.com/math-he...tml#post119542

    to see what needs to be done for the first. The details are slightly different but you should get the idea.

    The second follows the same blueprint as the first, it's just a different pdf that gets integrated and the lower integral terminal will be 0.

    Have another go at these two. Post at this thread if you struggle and I or someone else will work through a few more of the details.


    By the way, another thread of related interest: http://www.mathhelpforum.com/math-he...166-stats.html
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