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6) An urn contains eight green balls and nine white balls. A sample of four balls is selected at random from the urn.

(a) Find the probability that the four balls have the same color.

(b) Find the probability that the sample contains more green balls than white balls.

(a) The probability that the four balls have the same color is

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LammettHash

There is a total of 17 balls in the urn, and there are [tex]\dbinom{17}4=\dfrac{17!}{4!(17-4)!}=2380[/tex] ways of selecting 4 balls from it.

a.

[tex]P(\text{all green OR all white})=P(\text{all green})+P(\text{all white})=\dfrac{\binom84+\binom94}{\binom{17}4}=\boxed{\dfrac7{85}}[/tex]

b.

[tex]P(\text{3 green, 1 white OR all green})=P(\text{3 green, 1 white})+P(\text{all green})=\dfrac{\binom83\binom91+\binom84}{\binom{17}4}=\boxed{\dfrac{41}{170}}[/tex]

Using the hypergeometric distribution, it is found that there is a:

a) 0.0893 = 8.93% probability that the four balls have the same color.

b) 0.2412 = 24.12% probability that the sample contains more green balls than white balls.

The balls are chosen without replacement, which means that the hypergeometric distribution is used to solve this question.

Hypergeometric distribution:

[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:

  • The urn contains 8 + 9 = 17 balls, thus [tex]N = 17[/tex].
  • Eight of those are green, thus [tex]k = 8[/tex]
  • 4 are going to be selected, thus [tex]n = 4[/tex]

Item a:

Same color is none green, all white, which is P(X = 0), or all green, which is P(X = 4). Then:

[tex]p = P(X = 0) + P(X = 4)[/tex]

In which:

[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 = 0) = h(0,17,4,8) = \frac{C_{8,4}C_{9,0}}{C_{17,4}} = 0.0529[/tex]

[tex]P(X = 4) = h(4,17,4,8) = \frac{C_{8,0}C_{9,4}}{C_{17,4}} = 0.0294[/tex]

[tex]p = P(X = 0) + P(X = 4) = 0.0529 + 0.0294 = 0.0893[/tex]

0.0893 = 8.93% probability that the four balls have the same color.

Item b:

This is the probability of at least 3 green balls, thus:

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

In which:

[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 = 3) = h(3,17,4,8) = \frac{C_{8,1}C_{9,3}}{C_{17,4}} = 0.2118[/tex]

[tex]P(X = 4) = h(4,17,4,8) = \frac{C_{8,0}C_{9,4}}{C_{17,4}} = 0.0294[/tex]

Then:

[tex]P(X \geq 3) = P(X = 3) + P(X = 4) = 0.2118 + 0.0294 = 0.2412[/tex]

0.2412 = 24.12% probability that the sample contains more green balls than white balls.

A similar problem is given at https://brainly.com/question/24826394

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