In a certain lottery 5 numbers are drawn without replacement out of the set {1,2,...,90} randomly each week. Let X be the number of different numbers drawn during the first four weeks of the year. (For example, if the draws are {1,2,3,4,5}, {2,3,4,5,6}, {3,4,5,6,7} and {4,5,6,7,8} then X=8.) Find E(X).
My work:
Your approach seems good. Basically, you want to calculate $P(I_i = 1)$. For a given round, there are $5$ numbers drawn uniformly at random (uar) without replacement. So the probability of a certain number not being drawn in a given round is $$ p_5 = \frac{89}{90} \cdot \frac{88}{89} \cdot \frac{87}{88} \cdot \frac{86}{87} \cdot \frac{85}{86} = \frac{85}{90} = 1 - \frac{5}{90}. $$ (The subscript $5$ denotes that we draw $5$ numbers; in general it's a similar product of length $k$ if there are $k$ numbers drawn in one round.) Why is this? Let's assume that we don't want to hit number $90$. Draw the first one: what's the probability that it's not $90$? Exactly $89/90$. Now draw the second: probability $88/89$. Etc, etc.
Note that different rounds are independent, so the probability that a given number doesn't appear in $4$ weeks is $p_5^4$. (You can replace the power $4$ with a general $w$ number of weeks.)
Hence the probability that the number does appear is $1 - p_5^4$.
Now, here's the beauty of expectation. You might be thinking "but there are loads of correlations, so merely calculating the marginal for a single number isn't enough". But for expectation, it is! Indeed, $$ \textstyle E(X) = E\bigl( \sum_{i=1}^{90} 1(I_i = 1) \bigr), $$ where $1(\cdot)$ is the indicator function, and hence $$ \textstyle E(X) = \sum_{i=1}^{90} P(I_i = 1) = 90 \cdot (1 - p_5^4), $$ noting that each specific number $i$ has the same probability has a different number $j$.
Basically, the only part where you went wrong was calculating $p_5$ as a sum, rather than a product. It's a product, because you're saying "for all of the draws, the number does not come up".