Recall the example from tutorial where we had
X = # of heads after two coins are flipped.

The probability that there is exactly one head facing up when
the two coins land is P(X = 1) = 1/2.
The sample space is { HH, HT, TH, TT } and each has probability
1/4 of occurring.

Hence E[X] is 1/4 x 2 + 1/4 x 1 + 1/4 x 1 + 1/4 x 0 = 1.

In tutorial we went over why in an array A = A[1], A[2], ..., A[n],
which is equally likely to contain any permutation of {1,..., n},
the probability that A[i] is the largest amongst the elements A[1], A[2], ..., A[i] is just 1/i.

After class however, some of you mentioned that although you 
understood this, you were unclear why the expected value E[Xi] of
the random variable Xi is 1/i,
where Xi = 1 if A[i] is the largest element in A[1], ... , A[i], and 0 otherwise.

To see this we can simply invoke the definition of 
the expectation of a random variable:

E[Xi] = sum of x * P(Xi = x) where x takes on all values in the range of Xi, which is just {0 , 1}.
So E[Xi] = 0*P(Xi = 0) + 1*P(Xi = 1) = P(Xi = 1) = 1/i.

Then, as shown in tutorial:
E[total_number_of_assignments]
= E[X1 + ... + Xn]
= E[X1] + ... + E[Xn]  (by linearity of expectation)
= 1 + 1/2 + ... + 1/n.

Recall from tutorial that Xi is a random variable and hence is a function from the sample space (the set of permutations of {1,...,n})
to {0,1}.
Let's look at the special case where n=3 and i=2. We have that

X2(1,2,3) = 1
X2(1,3,2) = 1
X2(2,1,3) = 0
X2(2,3,1) = 1
X2(3,1,2) = 0
X2(3,2,1) = 0

Note that X2 = 1 in exactly 1/2 of the cases so P(X2 = 1) = 1/2 = 1/i.