g An urn contains 150 white balls and 50 black balls. Four balls are drawn at random one at a time. Determine the probability that there are 2 black balls and 2 white balls in the sample if .50 Each ball is replaced before the next one is drawn (sampling with replacement). .624 Each ball is not replaced (sampling without replacement)

Answer:

With replacement, 0.2109 = 21.09% probability that there are 2 black balls and 2 white balls in the sample.

Without replacement, 0.2116 = 21.16% probability that there are 2 black balls and 2 white balls in the sample.

Step-by-step explanation:

For sampling with replacement, we use the binomial distribution. Without replacement, we use the hypergeometric distribution.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.

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

And p is the probability of X happening.

Hypergeometric distribution:

The probability of x sucesses is given by the following formula:

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

In which:

x is the number of sucesses.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

[tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.

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

Sampling with replacement:

I consider a success choosing a black ball, so [tex]p = \frac{50}{150+50} = \frac{50}{200} = 0.25[/tex]

We want 2 black balls and 2 white, 2 + 2 = 4, so [tex]n = 4[/tex], and we want P(X = 2).

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 2) = C_{4,2}.(0.25)^{2}.(0.75)^{2} = 0.2109[/tex]

With replacement, 0.2109 = 21.09% probability that there are 2 black balls and 2 white balls in the sample.

Sampling without replacement:

150 + 50 = 200 total balls, so [tex]N = 200[/tex]

Sample of 4, so [tex]n = 4[/tex]

50 are black, so [tex]k = 50[/tex]

We want P(X = 2).

[tex]P(X = 2) = h(2,200,4,50) = \frac{C_{50,2}*C_{150,2}}{C_{200,4}} = 0.2116[/tex]

Without replacement, 0.2116 = 21.16% probability that there are 2 black balls and 2 white balls in the sample.

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