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

opatovaL

opatovaL

Answered question

2021-09-13

Suppose that you buy a lottery ticket containing k distinct numbers from among {1,2,...,n},1kn. 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?

Answer & Explanation

stuth1

stuth1

Skilled2021-09-14Added 97 answers

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)=n!r!(nr)! . 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 nk,then C(nk,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)=#of favorable outcomes#of possible outcomes=C(nk,k)C(n,k)=(nk)!/k!(n2k)!)n!/k!(nk)!=[(nk)!]2n!(n2k)! The sum of the probability of an event and the probability of the complementary ebent needs to be equal to 1 (P(E)+P(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)=1P(no winning number)1[(nk)!]2n!(n2k)!

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