In a lottery game, the host picks $6$ numbers out of $49$ numbers ranging consecutively from $1$ to $49$. After the six numbers are drawn, an additional number will be picked from the remaining numbers. What is the probability of you matching $3$ winning numbers and additional number?
My answer is:
$$\frac{\binom63 \cdot \binom11 \cdot \binom{42}2}{\binom{49}6 \cdot\binom{43}1}$$
My logic is that out $6$ winning cards, pick $3$ cards. Then for the one additional winning number, choose it. Then out of the remaining $42$ cards I pick the remaining $2$ cards. Can someone explain to me why this method does not work?
The correct answer is: $$\frac{\binom63 \cdot\binom{43}3}{\binom{49}6} \cdot \frac3{43}$$
Your error is in your "43C1" in your denominator. If you exclude it then the two answers would be the same
There are 49C6 i.e. ${49 \choose 6}$ ways of you making your $6$ choices from $49$. The fact that the lottery draw chooses a seventh bonus number does not affect that, as you do not choose a seventh number yourself.