College Algebra, Chapter 10, 10.3, Section 10.3, Problem 52

In the 6/49 lottery game a player selects six numbers from 1 to 49. What is the probability of selecting at least five of the six winning numbers?


$\begin{equation} \begin{aligned} P(\text{selecting at least 5 of the 6 winning numbers}) =& P(\text{getting exactly 5 of the 6 winning numbers}) \\ \\ =& \frac{C(6,5) C(43,1) }{C(49,6)} + \frac{C(6,6) C(43,0)}{C(49,6)} \end{aligned} \end{equation}$



One must divide the number of combinations producing the given result by the total number of possible combinations. So $49 C 6 = 13,983,816$. The numerator equates to the number of ways one ca select the winning numbers multiplied by the number of ways one can select the losing numbers. For a score of $n$, there are $6 Cn$ ways of selecting $n$ winning numbers from the $6$ winning numbers. This means that there are $6-n$ losing numbers, which are chosen from the $43$ losing numbers in $43 C(6-n)$ ways. So the probability of selecting at least 5 of the 6 winning numbers is $\displaystyle \frac{37}{1997688} = 0.0000185$.