0
$\begingroup$

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:

enter image description here

$\endgroup$
1
  • $\begingroup$ 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
1
$\begingroup$

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".

$\endgroup$
10
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.