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In Canada's national 6-49 lottery, a ticket has 6 numbers each from 1 to 49, with no repeats. Find the probability of matching exactly 4 of the 6 winning numbers if the winning numbers are all randomly chosen.

The answer is $\frac{\binom{6}{4}\cdot\binom{43}{2}}{\binom{49}{6}}$, but I don't understand why that is the answer and not $\frac{\binom{6}{4}\cdot\binom{45}{2}}{\binom{49}{6}}$, since if you match 4 numbers than there are $\binom{49-4}{2}=\binom{45}{2}$ ways to choose the last 2 numbers.

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1 Answer 1

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If there are 6 winning numbers, then you want to take away the possibility of choosing any of them when you choose the last 2 numbers. So we take away all 6 winning numbers from 49 so we get

$$\frac{{6 \choose 4}{49-6 \choose 2}}{49\choose 6}=\frac{{6 \choose 4}{43 \choose 2}}{49\choose 6}$$

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