Bingo Probability Problem

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 $pmf(Y)=cmf(Y)-cmf(Y-1)$

(a)

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

$Y=\min (X_1,\dots , X_{183})$ 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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