PROB_Q3

Prob Quiz 3.pdf

Problem 2

For a fixed $k$, the probability of each elementary event with $k$ red draws is equal.

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Problem 3

Part a

The support of ${\mathcal{B}}_{20}^{25}$ is $\{20,21,22,23,24,25\}$. The support of ${\mathcal{M}}_{20}^{25}$ is $\{10,11,12,13,14,15\}$. Thus, $P({\mathcal{B}}_{20}^{25}={\mathcal{M}}_{20}^{25})=0$.

Part b

The support for ${\mathcal{M}}_3^5$ is $\{2,3,4\}$ and the support for ${\mathcal{A}}_3^5$ is $\{1,2,3\}$.

$\omega \in \Omega$	${\mathcal{M}}_3^5$	${\mathcal{A}}_3^5$	${\mathcal{B}}_3^5$	${\mu }_3^5$
$\{1,2,3\}$	2	1	3	2
$\{1,2,4\}$	2	1	4	7/3
$\{1,2,5\}$	2	1	5	8/3
$\{1,3,4\}$	3	1	4	8/3
$\{1,4,5\}$	4	1	5	10/3
$\{1,3,5\}$	3	1	5	3
$\{2,3,4\}$	3	2	4	3
$\{2,3,5\}$	3	2	5	10/3
$\{2,4,5\}$	4	2	5	11/3
$\{3,4,5\}$	4	3	5	4
$P(\mathcal{A}=1\	\ \mathcal{M}=2)=1$
$P(\mathcal{A}=1\	\ \mathcal{M}=3)=\frac{1}{2}$
$P(\mathcal{A}=1\	\ \mathcal{M}=4)=\frac{1}{3}$

$P(\mathcal{A}=2|\mathcal{M}=2)=0$
$P(\mathcal{A}=2|\mathcal{M}=3)=\frac{1}{2}$
$P(\mathcal{A}=2|\mathcal{M}=4)=\frac{1}{3}$

$P(\mathcal{A}=3|\mathcal{M}=2)=0$
$P(\mathcal{A}=3|\mathcal{M}=3)=0$
$P(\mathcal{A}=3|\mathcal{M}=4)=\frac{1}{3}$

$\mathcal{M}$ and $\mathcal{B}$ are clearly not independent, since $P(\{\mathcal{M}=2\}\cap \{\mathcal{B}=3\})=\frac{1}{10}\ne P(\mathcal{M}=2)P(\mathcal{B}=3)=\frac{3}{100}$.

Part c

${\mu }_3^n$ takes it minimum value on $\{1,2,3\}$, which is $2$. It takes its maximum value on $\{n-2,n-1,n\}$, which is $n-1$. Thus, the smallest interval containing the support of ${\mu }_3^n$ is $[2,n-1]$.

Part d

From the table, $P(\mu =\mathcal{M})=\frac{4}{10}$.

Part e

$E(\mathcal{A})=1\left(\frac{6}{10}\right)+2\left(\frac{3}{10}\right)+3\left(\frac{1}{10}\right)=1.5$
$E(\mathcal{B})=3\left(\frac{1}{10}\right)+4\left(\frac{3}{10}\right)+5\left(\frac{6}{10}\right)=4.5$
$E(\mathcal{M})=2\left(\frac{3}{10}\right)+3\left(\frac{4}{10}\right)+4\left(\frac{3}{10}\right)=3$.

Part f

We have

$${\mu }_3^5=\frac{{\mathcal{A}}_3^5+{\mathcal{B}}_3^5+{\mathcal{M}}_3^5}{3}.$$

So,

$$\begin{array}{rl}E({\mu }_3^5)& =\frac{1}{3}(E({\mathcal{A}}_3^5)+E(B_3^5)+E(M_3^5))\\ & =3.\end{array}$$