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For this, I took n=2 which makes the set: {1,2,3,4}

Set will contain: {C1,C2,B1,B2}

X = 1 if the position of first cider bottle is 1

P(X=1) = 6/24 = 1/4

E(X) = 2 * 1/4 = 1/2

The general form will be: n*1/2n = 1/n.

This is my attempt, I'm not sure if I'm correct on this.

For this question:

You roll a fair die repeatedly and independently until the result is an even number. Defi ne the random variables X = the number of times you roll the die and Y = the result of the last roll. For example, if the results of the rolls are 5; 1; 3; 3; 5; 2, then X = 6 and Y = 2. Prove that the random variables X and Y are independent.

I defined X = 1 if number of times roll die is 1 time

and Y =1 if result of last roll is even

So, Pr(X) = 3/6 = 1/2 = Pr(Y)

Pr(X and Y) = 1/2

This gives me 1/2 = 1/4 which is not independent but the question is asking to prove independence

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closed as unclear what you're asking by Did, NCh, Leucippus, Tianlalu, KReiser Dec 19 '18 at 7:10

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    $\begingroup$ Please don't ask multiple unrelated questions in one post. $\endgroup$ – Bungo Dec 17 '18 at 6:22
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$\{X=1\}$ is the event that the first bottle is a cider bottle.

Probability on that: $P(\text{first cider})=\frac{n}{n+2}$

$\{X=2\}$ is the event that the first bottle contains beer and the second bottle contains cider.

Probability on that: $P(\text{first beer})P(\text{second cider}\mid\text{ first beer})=\frac2{n+2}\frac{n}{n+1}$.

$\{X=3\}$ is the event that the first bottle contains beer and the second bottle contains beer.

Probability on that: $P(\text{first beer})P(\text{second beer}\mid\text{ first beer})=\frac2{n+2}\frac{1}{n+1}$.

Now we are ready to find:$$\mathbb EX=P(X=1)+2P(X=2)+3P(X=3)=\frac{n}{n+2}+2\frac2{n+2}\frac{n}{n+1}+3\frac2{n+2}\frac1{n+1}=\frac{n+3}{n+1}$$


There are $2$ bottles that have index $1$ so that $P(Y=1)=\frac2{n+2}$.

$\{X=1,Y=1\}$ is the event that the first bottle is the cider bottle with index $1$.

Probability on that: $P(X=1,Y=1)=\frac1{n+2}$.

So a necessary condition for independence is: $$\frac{n}{n+2}\frac2{n+2}=P(X=1)P(Y=1)=P(X=1,Y=1)=\frac1{n+2}$$

leading to $n=2$.

So we conclude that there is no independence if $n>3$ and there might be independence if $n=2$. To verify we must check for that case whether $P(X=i)(Y=j)=P(X=i,Y=j)$ for $i,j\in\{1,2\}$.

I leave that to you.

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For the first part of the first problem, $X$ can take only values $1,2,$ and $3$.

For $X = 1$

$C_i ---------------$ in the first place and the rest can be filled with $2B_i$s and $(n-1) C_i$s.

For X =2

$B_i C_i ---------------$ one of the $B_i$s in the first place, $C_i$ in the second place and the rest can be filled with the remaining $B_i$s and $(n-1) C_i$s

For X = 3

$B_iB_i ----------------- $ Both the $B_i$s should occupy the first two places and the rest can be filled with the remaining $C_i$s.

Number of ways X = 1 can happen is = ${n\choose1} (n+1)!$

Number of ways X = 2 can happen is = ${2\choose1}{n\choose1} n!$

Number of ways X = 3 can happen is = ${2\choose1} n!$

Total number of ways =$(n+2)!$

Sanity check to see if $P(X=1)+P(X=2)+P(X=3) = 1$

$$\frac{(n(n+1)! + 2nn! + 2n!)}{(n+2)!} = 1$$

Thus the expected value $$ E(X) = \frac{1.{n\choose1}(n+1)!+2.{2\choose1}{n\choose1} n!+3.{2\choose1} n!}{(n+2)!} = \frac{n+3}{n+1}$$

Second Part

For $Y = 1$

$(B_1) ---------------$ $B_1$in the first place and the rest can be filled with the other $B_2$s and $(n)C_i$s to total of (n+1)! ways

$(C_1) ---------------$ $C_1$in the first place and the rest can be filled with the other $B_i$s and $(n-1)C_i$s to a total of (n+1)! ways.

Thus $P(Y=1) = \frac{(2(n+1)!)}{(n+2)!}$.

For Y =2

$-(B_1) ---------------$ The first place can be occupied with $(C_2-C_n)$ and $B_2$ and $B_1$in the second place and the rest can be filled with the other $B_2$s and $(n)C_i$s to total of ${n\choose1}n!$ ways

$-(C_1) ---------------$ The first place can be occupied with $(C_2-C_n)$ and $B_2$ and $C_1$in the second place and the rest can be filled with the other $B_2$s and $(n)C_i$s to total of ${n\choose1}n!$ ways

Thus$P(Y=2) = \frac{(2n) n!}{(n+2)!}$

and so on for (Y=n+1) for which the probability = $P(Y=n+1) = \frac{(2) n!}{(n+2)!}$

Thus $$E(Y) = \frac{2(n+1).n! \times 1 + 2n.n!\times 2 + 2(n-1)n!\times 3 +\cdots + 2(2).n!\times n+ 2(1).n!\times (n+1)}{(n+2)!} $$ $$= \frac{(n+3)}{3}$$

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