Question

An urn contains 4 red balls, 5 green balls and 3 yellow balls. An experiment consists of picking 4 balls simultaneously. What is the probability that you pick at least 3 green balls

Answer 1

Using the hypergeometric distribution, it is found that there is a 0.1515 = 15.15% probability that you pick at least 3 green balls.

The balls are chosen without replacement, hence the hypergeometric distribution is used.

What is the hypergeometric distribution formula?

The formula is:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}$

$C_{n,x} = \frac{n!}{x!(n-x)!}$

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.

In this problem:

• There are 4 + 5 + 3 = 12 balls, hence N = 12.

• 5 of the balls are green, hence k = 5.

• 4 balls will be picked, hence n = 4.

The probability that you pick at least 3 green balls is:

$P(X \geq 3) = P(X = 3) + P(X = 4)$

Hence:

$P(X = 3) = h(3,12,4,5) = \frac{C_{5,3}C_{7,1}}{C_{12,4}} = 0.1414$

$P(X = 4) = h(4,12,4,5) = \frac{C_{5,4}C_{7,0}}{C_{12,4}} = 0.0101$

Then:

$P(X \geq 3) = P(X = 3) + P(X = 4) = 0.1414 + 0.0101 = 0.1515$

0.1515 = 15.15% probability that you pick at least 3 green balls.

You can learn more about the hypergeometric distribution at brainly.com/question/4818951