A Bingo card has 25 squares with numbers on 24 of them, the center being a free square. The integers that are placed on the Bingo card are selected randomly and without replacement from 1 to 75, inclusive. When a game is called “cover-up" is played, balls numbered 1 to 75, inclusive, are selected randomly and without replacement until a player covers each of the numbers on a card. Let X equal the number of balls that must be drawn to cover all the numbers on a single card.

(a) Find p.m.f. of X

I got (24/X)((51 choose (x-24))/(75 choose x)) and I'm pretty certain it's right.

(b) Find the mean and variance of X

mean = 72.96 variance = 5.725

(c) If there are 183 people playing the game together, assume indepedence, and let Y be number of balls that must be drawn to cover all the numbers on a single card of the winner (the first person(s) to have his card coverd), find p.m.f. of Y.

I have no idea how to find the pmf of Y

(d) Find the Mean and Variance of Y.

I'm pretty sure I can get it after I find the pmf of Y.

  • $\begingroup$ Can you find the CDF? to do this, you can think of the minimum of all X's, which is itself a random variable $\endgroup$
    – M Turgeon
    Nov 18 '13 at 0:56
  • $\begingroup$ How would I integrate the "((51 choose (x-24))/(75 choose x))"? Convert it into factorial form? $\endgroup$
    – Jesus
    Nov 18 '13 at 1:27
  • $\begingroup$ You don't need to integrate, just take the sum. The support of $Y$ is clearly finite. But they're may be a simpler way of doing this than taking the product of all marginal CDFs. $\endgroup$
    – M Turgeon
    Nov 18 '13 at 3:50
  • $\begingroup$ But the sum of the pmf of X is 1, how can I use that to find the pmf of Y? $\endgroup$
    – Jesus
    Nov 18 '13 at 4:17
  • $\begingroup$ Under a, 24/X looks like it should be ${24 \choose X}$. This cannot be correct, because if $X \lt 24$ the probability has to be zero. Also, please do not mix cases. $x \neq X$ You use $x$ twice here. $\endgroup$ Jan 28 '14 at 3:49

For a, it is easier to find the cmf of $X$. If you have drawn $X$ balls, the chance you have covered the card is simply $p(X)=\frac {X \choose 24}{75 \choose X}$ Now $pmf(X)=p(X)-p(X-1)$

For c, the chance that nobody has won after $Y$ balls is $1-(1-p(Y))^{183}$ This is the cmf of $Y$, so $pmf(Y)=cmf(Y)-cmf(Y-1)$



If $m\ge 24$ balls have been drawn, the probability that exactly $k$ of your numbers are covered is:

${24 \choose k}\cdot {51 \choose m-k}$ out of a possible ${75 \choose m}$ ways to draw

Thus $P(X\le m)=\frac{ {24 \choose 24} \cdot {51 \choose m-24}}{{75\choose m}}=\frac{{51 \choose m-24}}{{75\choose m}}$

Now: $P(X=m)=P(X\le m)-P(X\le m-1)=\frac{{51 \choose m-24}}{{75 \choose m}}-\frac{{51 \choose m-25}}{{75 \choose m-1}}=\frac{{m-1 \choose 23}}{{75 \choose 24}}$

(b) $E[X]=\sum_{m=24}^{75} mP(X=m)=\frac{24}{{75\choose 24}}\sum_{m=24}^{75} {m\choose 24}=\frac{24}{{75\choose 24}}\cdot {76 \choose 25}=\frac{24\cdot 76}{25}=72.96$ $E[X^2]=\sum_{m=24}^{75} (m^2+m-m)P(X=m)=\frac{24\cdot 25}{{75 \choose 24}}\sum_{m=24}^{75} {m+1 \choose 25}-E[X]$ $=\frac{24\cdot 25}{{75 \choose 24}}\cdot {77 \choose 26}-E[X]=\frac{24 \cdot 77 \cdot 76}{26}-\frac{24\cdot 76}{25}=\frac{24\cdot 76\cdot (76\cdot 25-1)}{25\cdot 26}$

So $Var[X]=E[X^2]-(E[X])^2=\frac{24\cdot 76 \cdot 51}{25^2\cdot 26}\approx 5.725$

(c) Let $X_1,X_2,\dots, X_{183}$ be the independent copies of $X$ for each individual player

$Y=\min (X_1,\dots , X_{183})$ so $P(Y\ge m) = P(X_1,\dots X_{183}\ge m)=\left[ P(X\ge m)\right]^{183}$ $=\left[1-P(X\le m-1)\right]^{183}=\left[1-\frac{{51\choose m-25}}{{75\choose m-1}}\right]^{183}$

Thus we can get $P(Y=m)=P(Y\ge m)-P(Y\ge m+1)=\left[1-\frac{{51\choose m-25}}{{75\choose m-1}}\right]^{183}-\left[1-\frac{{51\choose m-24}}{{75\choose m}}\right]^{183}$

This is signficantly less pretty to manipulate, perhaps you could use the approximation $(1-\alpha)^n \approx 1-n\alpha$ to get the sums to go through the same motions?

for (d) $E[Y]=\sum_{m=24}^{75} m\left(\left[1-\frac{{51\choose m-25}}{{75\choose m-1}}\right]^{183}-\left[1-\frac{{51\choose m-24}}{{75\choose m}}\right]^{183}\right)\approx 62.1453$

and $E[Y^2]=\sum_{m=24}^{75} m^2 \left(\left[1-\frac{{51\choose m-25}}{{75\choose m-1}}\right]^{183}-\left[1-\frac{{51\choose m-24}}{{75\choose m}}\right]^{183}\right)\approx 3868.619796$

so $Var[Y]=E[Y^2]-(E[Y])^2\approx 3868.619796 - 62.1453^2=6.5814$


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