Question

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

Answer 1

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

a.

`P(\text{all green OR all white})=P(\text{all green})+P(\text{all white})=\frac{\binom84+\binom94}{\binom{17}4}=\boxed{\frac7{85}}`

b.

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