n cample from a small finite population without replacement, the binomial distribution should not be used because the events are not independent. If sampling is done without replacement and the outcomes belong to one of two types, we n Use the hypergeometric distribution. If a population has A objects of one type, while the remaining B objects are of the other type, and if n objects are sampled without replacement, then the probability of getting x objects of type A and n-x iects of type B under the hypergeometric distribution is given by the following formula. In a lottery game, a bettor selects four numbers from 1 to 52 (without repetition), and a winning four-number combination is later randomly selected, Find ne probabilities of getting exactly two winning numbers with one ticket. (Hint: Use A = 4, B= 48, n= 4, and x=2.) BI A P(x) = (A+B) (A- x)lx (B-n+x)(n-x) (A+B-n)!n! P(2)-1 (Round to four decimal places as needed.)

Algebra and Trigonometry (MindTap Course List)
4th Edition
ISBN:9781305071742
Author:James Stewart, Lothar Redlin, Saleem Watson
Publisher:James Stewart, Lothar Redlin, Saleem Watson
Chapter14: Counting And Probability
Section14.2: Probability
Problem 3E: The conditional probability of E given that F occurs is P(EF)=___________. So in rolling a die the...
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f we sample from a small finite population without replacement, the binomial distribution should not be used because the events are not independent. If sampling is done without replacement and the outcomes belong to one of two tynes we
can use the hypergeometric distribution. If a population has A objects of one type, while the remaining B objects are of the other type, and if n objects are sampled without replacement, then the probability of getting x objects of type A and n-x
objects of type B under the hypergeometric distribution is given by the following formula. In a lottery game, a bettor selects four numbers from 1 to 52 (without repetition), and a winning four-number combination is later randomly selected, Find
the probabilities of getting exactly two winning numbers with one ticket. (Hint: Use A=4, B= 48, n =4, and x 2.)
A!
P(x)=
BI
(A+ B).
(A-x)lx! (B-n+x)l(n-x)! (A+B-n)!n!
P(2) = 1
(Round to four decimal places as needed.)
Textbook
StatCrunch
O Time Remaining: 00:33:47
Next
OCT
25
ckook
talele
tab
D
G
Transcribed Image Text:f we sample from a small finite population without replacement, the binomial distribution should not be used because the events are not independent. If sampling is done without replacement and the outcomes belong to one of two tynes we can use the hypergeometric distribution. If a population has A objects of one type, while the remaining B objects are of the other type, and if n objects are sampled without replacement, then the probability of getting x objects of type A and n-x objects of type B under the hypergeometric distribution is given by the following formula. In a lottery game, a bettor selects four numbers from 1 to 52 (without repetition), and a winning four-number combination is later randomly selected, Find the probabilities of getting exactly two winning numbers with one ticket. (Hint: Use A=4, B= 48, n =4, and x 2.) A! P(x)= BI (A+ B). (A-x)lx! (B-n+x)l(n-x)! (A+B-n)!n! P(2) = 1 (Round to four decimal places as needed.) Textbook StatCrunch O Time Remaining: 00:33:47 Next OCT 25 ckook talele tab D G
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