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)

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.

In which is the number of different combinations of x objects from a set of n elements, given by the following formula.

And p is the probability of X happening.

Hypergeometric distribution:

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

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.

is the number of different combinations of x objects from a set of n elements, given by the following formula.

Sampling with replacement:

I consider a success choosing a black ball, so

We want 2 black balls and 2 white, 2 + 2 = 4, so , and we want P(X = 2).

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

Sample of 4, so

50 are black, so

We want P(X = 2).

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

seems like d

step-by-step explanation: