Probability with Equally Likely Outcomes

Problem 105: An urn contains eight white balls and two green balls. A sample of three balls is selected at random.

(a) What is the probability that the sample contains only white balls.

(b) What is the probability that the sample contains at least one green ball?

Solution: The probability of an event, E, is given by

$\text{Pr(E)} = \frac{n(E)}{N}$

where $n(E)$ is the number of outcomes in the event E, and $N$ the number of outcomes in the sample space. That is,

(a) Regarding of the color of the balls, we can select three balls in $\binom{10}{3}$ ways. Thus this is the number of outcomes in the sample space. We let $E$ be the event that all three balls selected are white. Therefore, the number of possible outcomes in $E$ is given by $\binom{8}{3}$ . Hence,

$\text{Pr(E)} = \frac{\binom{8}{3}}{\binom{10}{3}} = \frac{56}{120}=\frac{7}{15}.$

(b) In this occasion, the number of outcomes in the sample space is still the same. Let $F$ be the event that the sample contains at least one green ball. The the number of outcomes in $F$ is given by

$\underbrace{\binom{2}{1} \cdot \binom{8}{2}}_{\text{number of samples containing one green ball}} + \underbrace{\binom{2}{2} \cdot \binom{8}{1}}_{\text{number of samples containing two green balls}} .$

That is,

$\text{Pr(F)} = \frac{\binom{2}{1} \cdot \binom{8}{2} + \binom{2}{2} \cdot \binom{8}{1}}{\binom{10}{3}}= \frac{64}{120}=\frac{8}{15}.$

One Problem at a Time

Probability with Equally Likely Outcomes

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