A lottery consist of selecting $6$ numbers from 1-53. What is the probability of getting exactly 4 winning numbers in one ticket.
I think the denominator would be $(53)^6$ because you have
$53(53)(53)(53)(53)(53)$ possible ways to chose.
The chance of winning is $1/53$ and chance of losing is $52/53$
But I am not sure if the order matters. I am not sure what to do.
Best Answer
There are $$\binom{53}{6}$$ ways to select six of the $53$ numbers.
If exactly four of the six winning numbers are selected, then two of the other $47$ numbers must be selected. Hence, there are $$\binom{6}{4}\binom{47}{2}$$ ways to select exactly four of the six winning numbers.
Consequently, the probability that exactly four of the six winning numbers will be selected is $$\frac{\dbinom{6}{4}\dbinom{47}{2}}{\dbinom{53}{6}}$$
Why is your attempt incorrect?
Let's look at your denominator first.
There are $53$ ways to choose the first number. Once it has been chosen, there are only $52$ ways to choose the second number. That leaves $51$ ways to choose the third number, $50$ ways to choose the fourth number, $49$ ways to choose the fifth number, and $48$ ways to choose the sixth number. Therefore, if you wish to take order into account, there are $$53 \cdot 52 \cdot 51 \cdot 50 \cdot 49 \cdot 48$$ ways to select six numbers.
There are six ways to choose one of the six winning numbers. Once you to do so, there are five ways to choose one of the remaining winning numbers. Once two winning numbers have been chosen, there are four ways to choose a third winning number. That leaves three ways to choose a fourth winning number. Similarly, there are $47$ ways to choose one of the non-winning numbers. Once it has been chosen, there are $46$ ways to choose a second non-winning number. Therefore, the number of ways of choosing four winning numbers and two non-winning numbers in that order is $$6 \cdot 5 \cdot 4 \cdot 3 \cdot 47 \cdot 46$$ However, the four winning numbers can be selected in any of the six positions. There are $\binom{6}{4}$ ways that the winning numbers could appear in exactly four of the six positions. Hence, when the order of selection is taken into account, the number of favorable cases is $$\binom{6}{4} \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 47 \cdot 46$$ Therefore, the probability of selecting exactly four of the six winning numbers is $$\frac{\binom{6}{4} \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 47 \cdot 46}{53 \cdot 52 \cdot 51 \cdot 50 \cdot 49 \cdot 48}$$ You should verify that this agrees with the answer given above.