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If 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 types,
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 54 (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 = 50, n = 4, and
x=2.)
P(x)=
A!
B!
(A + B)!
(A-x)!x! (B-n+x)!(n-x)! (A+B-n)!n!
+


P(2)=
(Round to four decimal places as needed.)

Respuesta :

Using the hypergeometric distribution, it is found that there is a 0.0232 = 2.32% probability of getting exactly two winning numbers with one ticket.

What is the hypergeometric distribution formula?

The formula is:

[tex]P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}[/tex]

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

The parameters are:

  • x is the number of successes.
  • N is the size of the population.
  • n is the size of the sample.
  • k is the total number of desired outcomes.

For this problem, the parameters are given as follows:

N =A + B = 54, k = 4, n = 4.

The probability of getting exactly two winning numbers with one ticket is P(X = 2), hence:

[tex]P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}[/tex]

[tex]P(X = 2) = h(2,54,4,4) = \frac{C_{4,2}C_{50,2}}{C_{54,4}} = 0.0232[/tex]

There is a 0.0232 = 2.32% probability of getting exactly two winning numbers with one ticket.

More can be learned about the hypergeometric distribution at https://brainly.com/question/24826394

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