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 $\displaystyle P(X \leq k) \approx 0,99$My solution:

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

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

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

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

Okay now I got $\displaystyle n = 7$. But to me it is a bit of a elementary approach to just keep changing the value for $\displaystyle n$ 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 $\displaystyle r = 3$. I was struggling with thinking of a value for $\displaystyle k$, but logic tells me it must be a multiple of 3?

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

$\displaystyle 1 - {3-1 \choose 3-1} (0,5)^3 (1-0,5)^{3-3} $ $\displaystyle - {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 $\displaystyle \lambda = 0,1$

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 $\displaystyle t$ such that the probability that that the lifetime is greater than $\displaystyle t$ is 0,1My solution:

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

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

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

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

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

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

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

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

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

$\displaystyle = P(U \leq v^2)$

$\displaystyle = f_{U} (v^2)$

$\displaystyle = v^2$

Differentiate.

$\displaystyle f_{V} (v) = 2v$

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

My solution:

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

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

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

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

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

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

Differentiate. (Assuming the mass is a constant)

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

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

My solution:

$\displaystyle A = \pi r ^2$

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

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

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

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

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

Differentiate.

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

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Thanks in advance