College Algebra, Chapter 10, 10.3, Section 10.3, Problem 64

In the 6/49 lottery game, a player selects six numbers from 1 to 49 and wins if he or she selects the winning six numbers. What is the probability of winning the lottery two times in a row?


$
\begin{equation}
\begin{aligned}

P(\text{winning the lottery two times in a row}) =& P(\text{getting the winning numbers}) \times P(\text{getting the winning numbers})
\\
\\
=& \frac{C(6,6) C(43,0)}{C(49,6)} \times \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 winning the lottery two times in a row is $\displaystyle \frac{1}{13,983,816} \times \frac{1}{13,983,816} = \frac{1}{1.955471099 \times 10^{14}}$