The point is very simple: The order of flips matters.
You are given four random variables $X_i, i = 1,2,3,4$, where each one stands for a toss of a coin, in ascending order of indices. You have to calculate the following quantity : $X = 8X_1 + 4X_2 + 2X_3 + X_4$,right? This can have values between $0$ and $15$.
For the first part: Now, how is $X_i$? Well, it is $0$ or $1$ with probability half, so the expectation value is just $E[X] = 8 \times 0.5 + 4 \times 0.5 + 2 \times 0.5+1 \times 0.5 = 7.5$, because the expectation of one toss is $0 \times 0.5 +1 \times 0.5 = 0.5$.
How about the mass function? Well, each number, from $0$ to $15$, has an equal probability of coming, because each one has a unique representation in binary, and the probability of heads and tails are the same. Hence, it is just $P(i) = \frac{1}{16}$, for $i=0,1, \ldots, 15$ (there are sixteen possibilities in total, each occurs equally probably).
In the second question: The expectation of each $X_i$ now is $0 \times \frac 13 + 1\times \frac 23 = \frac 23$, so then $E[X] = (8+4+2+1) \times\frac 23 = 10$.
Finally, here, every number has a unique binary expansion, but this time the heads and tails don't have the same probability, so the mass function changes, and it changes as follows: given the number of $1$s in the expansion, so many tails are required for the number to come, which happens with higher probability. So more the ones, more the probability.
So, we look at the binary expansions:
1) The numbers having no $1$s is only zero.
2) The numbers having precisely one $1$ are $2$,$4$,$8$,$1$.
3) The numbers having precisely two $1$s are $3$,$5$,$6$,$9$,$10$,$12$.
4) The numbers having precisely three ones are $7$,$11$,$13$,$14$.
5) Only $15$ has four ones.
Hence, we can write down the probabilities as follows:
1) For zero, the probability is $\frac{1}{3^4} = \frac{1}{81}$.
2) For the numbers with only one $1$, the probability is $\frac{1}{3^3} \times \frac{2}{3} = \frac{2}{81}$.
3) For the numbers having $2$ ones, the probability is $\frac{1}{3^2} \times \frac{2^2}{3^2} = \frac{4}{81}$.
4) For the numbers with three $1$s, the probability is $\frac{1}{3} \times \frac{2^3}{3^3} = \frac{8}{81}$.
5) For the numbers with all $1$s, the probability is $\frac{2^4}{3^4} = \frac{16}{81}$.
Hence, this is the probability distribution function in the end:
$$
P(x) = \begin{cases}
\frac{1}{81} & x=0 \\
\frac{2}{81} & x=1,2,4,8 \\
\frac{4}{81} & x= 3,5,6,9,10,12 \\
\frac{8}{81} & x = 7,11,13,14 \\
\frac{16}{81} & x = 15 \\
\end{cases}
$$
Best Answer
The first step is to determine what sample space we are working with. Since we have three coin flips we can have outcomes on the form $HHT$, $THH$, $TTT$, $\dots$ and so on, where $H=heads$ and $T=tails$. The total number of outcomes is $2^3 = 8$ and we can compute the probability of outcomes using independence. For instance $$P(THT) = \frac{1}{3}\cdot\frac{2}{3} \cdot\frac{1}{3} = \frac{2}{27}$$ Now let's suppose we want to calculate the joint probability of $X=2$ and $Y=0$. The only possible ways this can happen is $HTH$ and $HHT$, so $$P(X=2,Y=0) = P(HTH)+P(HHT) = \frac{2}{3}\cdot \frac{1}{3} \cdot \frac{2}{3} + \frac{2}{3}\cdot \frac{2}{3} \cdot \frac{1}{3}= \frac{8}{27}$$ And you can do the same for all other possible cases. The most convenient way would be to make a table of all possibilities:
\begin{array}{l|llll} X \quad \text{\\} \quad Y & 0 & 1 & 2 & 3 \\ \hline 0 & \text{Not Possible} & \text{Not Possible} & \text{Not Possible} & TTT \\ 1 & HTT & THT & TTH & \text{Not Possible} \\ 2 & HHT \text{ or } HTH & THH & \text{Not Possible} & \text{Not Possible}\\ 3 & HHH & \text{Not Possible} & \text{Not Possible} & \text{Not Possible} \end{array}
Now see if you can make a similar table where you compute all the probabilities. Note that events that are not possible will of course have probability 0.
ii) From here you can find the marginal distribution of $X$ by summing over the rows and the marginal distribution of $Y$ by summing columns.
iii) The last part of the question can be computed using the definition of conditional probability: $$P(X\leq 1 \: | \: Y\leq 1) = \frac{P(X\leq 1,Y\leq 1)}{P(Y\leq 1)}$$