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This is about a bingo game, where the numbers $1$ to $42$ are drawn one at a time in a random order without replacement. We are interested in when and how the six numbers in the fourth row in the bingo plate down below are drawn, that is, the numbers $24$, $37$, $4$, $11$, $6$ and $2$.

 30  33  16  10  42  38
  1  40  25  31   5   7
 17  41  19  27   8  13
 24  37   4  11   6   2
 20  22  28  39  23  12
  3  18  29  15  21  36
 26  35  14  34  32   9

a) What is the probability that all numbers in the fourth row are drawn in the following order $(24, 37, 4, 11, 6, 2)$, when the first six numbers have been drawn?

My answer: $$\frac{1}{42}\cdot \frac{1}{41}\cdot \frac{1}{40}\cdot \frac{1}{39}\cdot \frac{1}{38} \cdot \frac{1}{37}=\frac{1}{\frac{42!}{(42-6)!}}.$$

b) What is the probability that all numbers in the fourth row are drawn, when the first six numbers have been drawn? (The six numbers don't have to be drawn in the given order).

My answer: $$\frac{6}{42}\cdot \frac{5}{41}\cdot \frac{4}{40}\cdot \frac{3}{39}\cdot \frac{2}{38} \cdot \frac{1}{37}=\frac{1}{\binom{42}{6}}.$$

c) What is the probability that five of the six numbers in the fourth row have been drawn after having drawn $41$ numbers?

My answer: $$\frac{41}{42}\cdot \frac{1}{\binom{42}{5}}.$$

d) What is the probability that all numbers in the fourth row are drawn, when the first seven numbers have been drawn?

My answer: I am not sure. I think it's $$\frac{1}{\frac{42!}{(42-7)!}}\cdot\frac{1}{\binom{42}{6}}.$$

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Your answers to (a) and (b) look good, but it seems like (c) and (d) are a bit off.

Think about (c) this way: in order for exactly five of the six to be drawn after having drawn $41$ numbers, we need the last number drawn to be in the fourth row. Each number has an equal probability of being chosen last, so this number is merely $\frac{6}{42}$.

For (d), there are a total of $\binom{42}{7}$ ways to choose the first $7$ balls, and each of these ways occurs with equal probability. In how many of these ways do all numbers in the fourth row appear? Since there are six numbers that we know must appear, we have $42 - 6 = 36$ ways to pick $7$ numbers if we require that all numbers in the fourth row appear. Thus, we have a probability of $$ \frac{36}{\binom{42}{7}}$$ of this occurring.

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