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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

Sagot :

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:

[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.

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:

[tex]P(X \geq 3) = P(X = 3) + P(X = 4)[/tex]

Hence:

[tex]P(X = 3) = h(3,12,4,5) = \frac{C_{5,3}C_{7,1}}{C_{12,4}} = 0.1414[/tex]

[tex]P(X = 4) = h(4,12,4,5) = \frac{C_{5,4}C_{7,0}}{C_{12,4}} = 0.0101[/tex]

Then:

[tex]P(X \geq 3) = P(X = 3) + P(X = 4) = 0.1414 + 0.0101 = 0.1515[/tex]

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

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