Question # Suppose that you buy a lottery ticket containing k distinct numbers from among $\left\{1,2,...,n\right\}, 1\leq k \leq n$.

Probability and combinatorics
ANSWERED Suppose that you buy a lottery ticket containing k distinct numbers from among $\left\{1,2,...,n\right\}, 1\leq k \leq n$. To determine the winning tickets, k balls are randomly drawn without replacement from a bin containing n balls numbered 1, 2, . . . , n. What is the probability that at least one of the numbers on your lottery ticket is among those drawn from the bin? An r-combination of a set of n elements is a subset that contains r of the n elements. The number of r-combinations of a set of n distinct objects is $C(n,r)=\frac{n!}{r!(n-r)!}$ . There are C(n,k) ways to select k of the n numbers to be in the bin. #of possible outcomes=C(n,k) When we select none of the k winning numbers, then there are C(n-k,k) ways to select the k numbers from the remaining n-k numbers. Note: $When\ n-k, then\ C(n-k,k)=0$ #of favorable outcomes=C(n-k,k) The probability is the number of favorable outcomes divided by the number of possible outcomes: $P(no\ winnning\ numbers)=\frac{\#of\ favorable\ outcomes}{\#of\ possible\ outcomes} \\ =\frac{C(n-k,k)}{C(n,k)} \\ =\frac{(n-k)!/k!(n-2k)!)}{n!/k!(n-k)!} \\ =\frac{[(n-k)!]^{2}}{n!(n-2k)!} \\$ The sum of the probability of an event and the probability of the complementary ebent needs to be equal to 1 $(P(E)+P(\overline{E})=1)$, which implies that the probability of the complementary enevt os 1 decreased by the probability of the event $P(at\ least\ one\ winning\ number)=1-P(no\ winning\ number) \\ 1-\frac{[(n-k)!]^{2}}{n!(n-2k)!}$