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:

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  • $\begingroup$ I am confused about how you can come up the P(Ii=1) where there is without replacement part(days) and with the replacement part(week). $\endgroup$ – Michelle Feb 26 '19 at 20:57

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".

  • $\begingroup$ Maybe I didn't understand the problem quite right. Why isn't $p_5$ just equal to $85/90$? $\endgroup$ – Brian Tung Feb 26 '19 at 21:10
  • $\begingroup$ Thank you! Just one question, so since if they repeat, Ii=0. $\endgroup$ – Michelle Feb 26 '19 at 21:14
  • $\begingroup$ So would I think as P(Ii=1)=P(i showed up at least twice during the 4 weeks)=1-P(i showed once)-P(i didn't show up)? $\endgroup$ – Michelle Feb 26 '19 at 21:15
  • $\begingroup$ @BrianTung The five numbers are chosen uniformly at random without replacement. 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 $88/89$... and whoops, I see exactly what you mean, I've been a fool and used copy+paste too recklessly! Let me correct that... $\endgroup$ – Sam OT Feb 26 '19 at 21:16
  • $\begingroup$ @Michelle I don't understand what you mean by "so since if they repeat, $I_i = 0$"? $I_i$ is simply the indicator function that the number shows up at some point. If it shows up 4 times, then $I_i = 1$; if once, then $I_i = 1$; only if it never shows up is $I_i = 0$. $\endgroup$ – Sam OT Feb 26 '19 at 21:19

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