# Thread: A couple of problems

1. ## A couple of problems

1.
Three urns (A,B,C) hold white, red and black beads.

white, red, black

A 10, 5, 10
B 8, 8, 4
C 5, 5, 5

You choose an urn at random and take two beads out of this urn (without putting anyone of them back).

You have a black and a white bead. What is the probability that you choose urn A?

2.
Decide the value of k so the function $f(x) = k \cdot x \cdot e^{\frac{-x^2}{2a^2}}$ $, x > 0 (\sigma > 0)$ becomes the density function for the random variable $X$.

Then calculate $P(X > \sigma)$. (Rayleigh distribution)

3.
On average, 5.2 hurricanes occur in a year. What is the probability three or less will occur this year?

We can assume that the number of hurricanes can be approximated with a Poisson random variable.

Thank you.

2. Originally Posted by Tau
1.
[snip]
2.
Decide the value of k so the function $f(x) = k \cdot x \cdot e^{\frac{-x^2}{2a^2}}$ $, x > 0 (\sigma > 0)$ becomes the density function for the random variable $X$.

Then calculate $P(X > \sigma)$. (Rayleigh distribution)

3.
On average, 5.2 hurricanes occur in a year. What is the probability three or less will occur this year?

We can assume that the number of hurricanes can be approximated with a Poisson random variable.

Thank you.
2. Equate the integral of the pdf to 1 and solve for k. What have you tried and where are you stuck?

3. Calculate Pr(X = 0) + Pr(X = 1) + Pr(X = 2) + Pr(X = 3). Each term is calcultaed from the pmf of the Poisson distribution and you're given the mean. What have yuo tried and where do you get stuck.

(Someone else can look at 1. - I'm tired of urn problems).

3. Hello, Tau!

1. Three urns (A,B,C) hold white, red and black beads.

. . $\begin{array}{c||c|c|c|}
& \text{white} & \text{red} & \text{black} \\ \hline\hline
\text{urn A} & 10 & 5 & 10 \\ \hline
\text{urn B} & 8 & 8 & 4 \\ \hline
\text{urn C} & 5 & 5 & 5 \\ \hline \end{array}$

You choose an urn at random and take two beads out of this urn
(without putting any one of them back).

You have a black and a white bead.
What is the probability that you chose urn $A$?

Bayes' Theorem: . $P(\text{urn A}\,|\,b\, \& w) \;=\;\frac{P(\text{urn A}\,\wedge\,b\,\&w)}{P(b\,\&w)}$ .[1]

Numerator

$P(\text{urn A}) \,=\,\frac{1}{3}$

$P(b\,\& w) \:=\:\frac{10}{25}\cdot\frac{10}{24} \:=\:\frac{1}{6}$

. . Hence: . $P(\text{urn A}\,\wedge\,b\,\&w) \:=\:\frac{1}{3}\cdot\frac{1}{6} \:=\:\frac{1}{18}$ .[2]

Denominator

$P(\text{urn A}\,\wedge\,b\,\&w) \;=\;\frac{1}{18}$

$P(\text{urn B}\,\wedge\,b\,\&w) \:=\:\frac{1}{3}\left(\frac{4}{20}\cdot\frac{8}{19 }\right) \:=\:\frac{8}{285}$

$P(\text{urn C}\,\wedge\,b\,\&w) \;=\;\frac{1}{3}\left(\frac{5}{15}\cdot\frac{5}{14 }\right) \:=\:\frac{5}{126}$

. . Hence: . $P(b\,\&w) \:=\:\frac{1}{18} + \frac{8}{285} + \frac{5}{126} \;=\;\frac{82}{665}$ .[3]

Substitute [2] and [3] into [1]:

. . $P(\text{urn A}\,|\,b\,\&w) \;=\;\frac{\dfrac{1}{18}}{\dfrac{82}{665}} \;=\;\frac{665}{1476}$

But check my work . . . please!
.