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The four musketeers decide to send a bouquet of roses to Snow White for
her Birthday. They toss a die each and contribute as many roses to the
bouquet as their die shows.

(i) What is the probability that at least two out of the four musketeers will
contribute different numbers of flowers?

For this part I assumed the probability of two number of flowers not matching is

$$frac{6}{6}times frac{5}{6}timesfrac{4}{6}timesfrac{3}{6}$$

For the first roll some number will be rolled so probability
is 1. For the second roll theres a 1/6 chance the roll will match the first so a 5/6 chance is won't match. For the third there's a 4/6 chance it won't match the 1^st or 2^nd roll and then with the 4^th roll there's a 3/6 chance it will match one of the others.
Similarly for the probability that 3 and 4 choose different numbers of flowers I get

$$frac{6}{6}times frac{5}{6}timesfrac{4}{6}timesfrac{3}{6}$$
and

$$frac{6}{6}times frac{5}{6}timesfrac{4}{6}timesfrac{3}{6}$$I think this isn't right. Do i instead need to work with 1 - probability of same number.

(ii) Let X be a random variable representing the maximal out of the four
contributions, and let Y stands for the minimal one. Calculate the value
of joint probability, PXY (3, 2), that X = 3 AND Y = 2.

For the number of possibilities that satisify this we are selecting 4 outcomes (4 dice rolls) and there are 2 possibilities for each outcome, either a 2 or a 3, so there would be 2⁴ possibilities and a sample space of 6⁴ so the probability is 2⁴/6⁴?

(iii) Calculate the expected difference between the maximal and the minimal contributions, E(X − Y ).

I'm not too sure about this one, but would it be, the probability of having a minimum of 2 would be 5⁴/6⁴ and the probability of the maximum being 3 is 3⁴/6⁴ so would he expected difference be

$$(2)(5^4/6^4)-(3)(3^4/6^4)$$

In part (ii), you also need to be sure that not all the rolls are all $2$ or $3$. As such, your answer should be

$$
(2/6)^4 - (1/6)^4 - (1/6)^4
$$

Since there is a $(1/6)^4$ chance of getting all 2, which is equal to chance of getting all 3.

This can also be done by multinational experiment.

$$
sum_{n=1}^3 frac{4!}{n!(4-n)!}(1/6)^4
$$

In part (iii), the dice rolls is uniform with a pmf of $1/6$ for $i = 1,2,...,6$. As such, the cdf is

$$
F(y) = y/6
$$

Notice that $P(Y_{max} < y)$ occurs only if all other rolls are less than $y$ also. Therefore

$$
P(Y_{max} leq y) = F(y)^4 = (y/6)^4
$$

As such, using $G(y)$ and $H(y)$ to simplify notation, we have

$$
G(y) = P(Y_{max} = y) = P(Y_{max} leq y) - P(Y_{max} leq (y-1)) = (y/6)^4 - ((y-1)/6)^4
$$

Similarly, since $P(Y_{min} geq y) = 1-(1-F(y))^4$, we have

$$
H(y) = P(Y_{min} = y) = P(Y_{min} leq y) - P(Y_{min} leq (y-1)) = ((7-y)/6)^4 - ((6-y)/6)^4
$$

Therefore, summing all possible values of y (1 through 6).

$$E(X-Y) = E(X) - E(Y) = sum yG(y) - sum yH(y) \= 5.24459876543 - 1.75540123457 = 3.48919753086
$$

For (i) You started out with the right approach, but the wrong question, although your last remark is correct. The opposite possibility is that all four have the same number which is $P=(frac{1}{6})^3$, so the probability you want is $1-(frac{1}{6})^3=0.99537037037037$

For (ii) You need to subtract the probabilities of all 3 or all 2 which $frac{2}{6^4}$.

For (iii) Let $Z=Y-X$, so you want $E(Z)=sum_{z=0}^5zP(Z=z)$. To get $P(Z=z)$, we have $P(Y-X=z)=P(X=y-z,Y=y)$, and we will need to sum over $y$. $P(X=y-z,Y=y)=P(X=y-z|Y=y)P(Y=y)$. $P(X=y-z|Y=y)={P(y-zle Xle y|Y=y)-P(y-z+1le Xle y|Y=y)=(frac{z+1}{y})^4-(frac{z}{y})^4}$.

$P(Y=y)=P(Yle y)-P(Yle y-1)=(frac{y}{6})^4-(frac{y-1}{6})^4$. Therefore $P(Z=z)=sum_{y=z+1}^6 [(frac{z+1}{y})^4-(frac{z}{y})^4][(frac{y}{6})^4-(frac{y-1}{6})^4]$

I'll leave it to others to do the arithmetic. Note that I have switch $X$ and $Y$. $Y$ is max, $X$ is min.

Hint on (iii).

By linearity of expectation: $mathbb E(X-Y)=mathbb EX-mathbb EY$.

What follows makes it more easy to find $mathbb EX$ and $mathbb EY$.

If $Z$ takes values in ${1,2,3,4,5,6}$ then: $$mathbb EZ=sum_{k=1}^6kP(Z=k)=sum_{k=1}^6sum_{r=1}^kP(Z=k)=sum_{r=1}^6sum_{k=r}^6P(Z=k)=sum_{r=1}^6P(Zgeq r)$$

Applying this on $X$ we find:$$mathbb EX=sum_{r=1}^6P(Xgeq r)=sum_{r=1}^6left(1-P(X<r)right)=6-sum_{r=1}^6P(X<r)$$where $P(X<r)$ is quite easy to find.

Also you apply this on $Y$ where $P(Ygeq r)$ is quite easy to find.