Let $X$ be the random variable that is defined as the smaller of the two numbers that appear when a pair of dice is rolled. For example, if you roll 2 and 5, then $X = 2$.
a) Determine the expected value and the standard deviation of $X$ for two fair dice whose results are independent of each other.
What I've got so far: $$s={1, 2, 3, 4, 5, 6}$$ $P=1/6$ for all outcomes
36 possibilities for all outcome considering a pair of two dice $(a,b)$. I need to find Expected value $E(X)$, and the Standard deviation $= \sqrt{\operatorname{Var}(X)}$ I've got $$E(X) = 1(1) + 3(2) + 5(3) + 7(4) + 9(5) + 11(6)=\frac{161}{36}$$
b) Someone inadvertently steps on one of the two dice, thereby flattening it so that from now on it can only show the numbers one and six (with equal probability). Determine the expected value (no standard deviation) of $X$ after this accident.
c) To make things worse, suddenly a mysterious mechanism comes into effect: The sum of the numbers of the two dice (the flat and the normal one) is never seven anymore, while all other possible outcomes, i.e., pairs of numbers, occur with equal probability. What is the new expected value of $X$?
$$E(X) = \frac{5(6) + 3(5)}{10}$$
d) In the final, mysterious state of the dice, are the numbers that appear on the flat die ($X_1$) and on the normal die ($X_2$) independent? Prove your answer.
For (a),
$$X=\min(X_1,X_2)$$ where $X_i$ is the outcome from each die.
So the possible outcomes are
\begin{array}{c|cccccc} &1&2&3&4&5&6\\ \hline 1&1&1&1&1&1&1\\ 2&1&2&2&2&2&2\\ 3&1&2&3&3&3&3\\ 4&1&2&3&4&4&4\\ 5&1&2&3&4&5&5\\ 6&1&2&3&4&5&6\\ \end{array}
So as you have done (but somehow with max instead), we have $$E(X)=\frac{1(11)+2(9)+3(7)+4(5)+5(3)+6(1)}{36}=\frac{91}{36}$$
and
$$E(X^2)=\frac{1^2(11)+2^2(9)+3^2(7)+4^2(5)+5^2(3)+6^2(1)}{36}=\frac{301}{36}$$
So $$\sigma=\sqrt{E(X^2)-E(X)^2}=\frac{\sqrt{2555}}{36}$$
For (b),
Now the possible outcomes become
\begin{array}{c|cccccc} &1&2&3&4&5&6\\ \hline 1&1&1&1&1&1&1\\ 6&1&2&3&4&5&6\\ \end{array}
So we have $$E(X)=\frac{1(7)+2(1)+3(1)+4(1)+5(1)+6(1)}{12}=\frac{9}{4}$$
For (c),
We simply cross out the last column
\begin{array}{c|cccccc} &1&2&3&4&5&6\\ \hline 1&1&1&1&1&1&\times\\ 1&\times&2&3&4&5&6\\ \end{array}
So we have $$E(X)=\frac{1(5)+2(1)+3(1)+4(1)+5(1)+6(1)}{10}=2.5$$
For (d),
We simply check if $$P(X_1=x_1)P(X_2=x_2)=P(X_1=x_1,X_2=x_2)$$
holds.
We know that the probability distribution is simply
\begin{array}{c|ccccc} &1&2&3&4&5\\ \hline 1&\frac1{10}&\frac1{10}&\frac1{10}&\frac1{10}&\frac1{10}\\ 6&\frac1{10}&\frac1{10}&\frac1{10}&\frac1{10}&\frac1{10}\\ \end{array}
while $$P(X_1=1)=P(X_1=6)=\frac12$$ and $$P(X_2=1)=P(X_2=2)=\ldots=P(X_2=6)=\frac16$$ are still in effect.
Very clearly,
$$P(X_1=1,X_2=6)=0\ne\frac1{12}=P(X_1=1)P(X_2=6)$$
Hence, $X_1,X_2$ are no longer independent.