Determine the Expected Value of Uniformly random elements of sets 
Answer is D
The way I attempted this was that for X = MAX(a,b), the random variable X is equivalent to the max value of a and b. So, from the 2 sets, the probability of getting k from set {1,2...100} is $\frac{k}{100}$ and for the second set we have to select $\frac{k-1}{100}$ so that it is less than the first one. 
E(X) would then just be the summation of $x*$$\frac{{k*}{(k-1)}}{100}$, which is option B. 
Don't understand how its D? Any explanations? 
 A: You can use the following:
$$E(X)=\sum_{k=1}^{100} P(X\ge k)$$
$$P(X\ge k)=1-P(X\le k-1)$$
$$P(X\le k)=P(a\le k\text{ and }b\le k)$$
A: Your general logic is fine, although you calculate the probability incorrectly. Note that, if $\max(a,b) = k$, then we have one of three cases:


*

*$a = k$ and $b < k$.

*$a < k$ and $b = k$.

*$a = b = k$
It should be readily apparent that these cases are disjoint, so to find the probability $P(X = k)$, we can add the probabilities of these cases.
Since $a,b$ are uniform, it's really a matter of counting how many ways this can happen. There's one way for $a = k$ to happen and $k-1$ ways for $b < k$, so multiplying these gives $k-1$ ways; similarly, there's also $k-1$ ways to get $a < k$ and $b = k$. Finally, there's exactly one way to get $a = b = k$. Adding these gives you $1 + 2(k - 1)$; we divide by $100^2$ as the total number of outcomes for $a,b$. This give the probability $P(X = k) = \frac{1 + 2(k-1)}{100^2}$, from which you can derive the expected value in the way you do above.
