The lottery in our state consists of two drawings.  First, a MegaBall is picked from among 27 numbered balls.  Second, five WinnerBalls are picked from among 44 numbered balls. To win the lottery, you must pick the MegaBall number correctly and also pick the numbers on the five WinnerBalls (but you don't need to get the order right for the WinnerBalls).  What is the probability that the ticket I hold has the winning numbers?
The probability that the MegaBall matches is $\dfrac{1}{27}$ . The probability that the 5 WinnerBalls match is $\dfrac{1}{\binom{44}{5}}$. So my chances of winning are $\left(\dfrac{1}{27}\right)\times\left(\dfrac{1}{\binom{44}{5}}\right) = \boxed{\dfrac{1}{29,\!322,\!216}}$.
\dfrac{1}{29322216}
Level 4
Counting & Probability
Problem: The lottery in our state consists of two drawings.  First, a MegaBall is picked from among 27 numbered balls.  Second, five WinnerBalls are picked from among 44 numbered balls. To win the lottery, you must pick the MegaBall number correctly and also pick the numbers on the five WinnerBalls (but you don't need to get the order right for the WinnerBalls).  What is the probability that the ticket I hold has the winning numbers?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To determine the probability of winning the lottery, we need to calculate the total number of possible outcomes for both the MegaBall and the WinnerBalls. 

First, consider the MegaBall. There are 27 possible numbers, and only one of these is the winning number. Therefore, the probability of selecting the correct MegaBall is 1 out of 27, or:

P(MegaBall) = 1/27.

Next, we analyze the WinnerBalls. We need to choose 5 numbers from a set of 44. The order of selection does not matter, so we use combinations to calculate the number of ways to choose 5 balls from 44. The formula for combinations is given by:

C(n, k) = n! / (k!(n-k)!),

where n is the total number of items to choose from, k is the number of items to choose, and "!" denotes factorial.

In this case, n = 44 and k = 5. Thus, we calculate:

C(44, 5) = 44! / (5!(44-5)!) = 44! / (5! * 39!) = (44 * 43 * 42 * 41 * 40) / (5 * 4 * 3 * 2 * 1).

Calculating this step-by-step:

1. Calculate the numerator: 44 * 43 = 1892.
2. 1892 * 42 = 79464.
3. 79464 * 41 = 3268044.
4. 3268044 * 40 = 130721760.

Now, calculate the denominator: 5! = 120.

Now, divide the numerator by the denominator:

C(44, 5) = 130721760 / 120 = 1089348.

This means there are 1,089,348 different combinations of WinnerBalls.

To find the total number of outcomes for the lottery, we multiply the number of outcomes for the MegaBall by the number of combinations for the WinnerBalls:

Total outcomes = 27 * 1089348 = 29322216.

Finally, since there is only one winning combination of numbers (the correct MegaBall and the correct set of WinnerBalls), the probability of holding a winning ticket is the reciprocal of the total number of outcomes:

P(winning ticket) = 1 / Total outcomes = 1 / 29322216.

Thus, the final answer is 1/29322216.<|END_OF_LATENT|> The final answer is $\dfrac{1}{29322216}$.