College Algebra, Chapter 10, Review Exercises, Section Review Exercises, Problem 24

If two balls are drawn from the jar in Exercise 23, find the probability of the given event.

a.) Both balls are red.

The probability that the first red ball is picked is $\displaystyle \frac{10}{15}$, then the probability that first and second red ball is picked is $\displaystyle \frac{10}{15} \times \frac{9}{14} = \frac{3}{7}$

b.) One ball is white and the other is red.

In this case, we must get the probability that the first ball picked is white and second is red or the probability that first ball picked is red and the second is white. Thus, we have

$\displaystyle \frac{5}{15} \times \frac{10}{14} + \frac{10}{15} \times \frac{5}{14} = \frac{10}{21}$

c.) At least one ball is red.

To solve this more easily, we can apply the complement to the probability that two white balls are picked. This gives

$\displaystyle 1 - \left( \frac{5}{15} \times \frac{4}{14} \right)= \frac{19}{21}$

d.) Both balls are red and even-numbered.

There are ten out of fifteen balls that are color red in which the numbers $0,2,4,6$ and $8$ are even. Thus, if the probability that first ball is red and even numbered is $\displaystyle \frac{5}{15}$. Then, the probability that the first and second ball is red and even numbered is $\displaystyle \frac{5}{15} \times \frac{4}{14} = \frac{2}{21}$

e.) Both balls are white and odd-numbered.

There are five out of fifteen balls that are color white in which the numbers $1$ and $3$ are odd. Thus, if the probability that first ball is white and odd numbered is $\displaystyle \frac{2}{15}$, then the probability that the first and second ball is white and odd numbered is $\displaystyle \frac{2}{15} \times \frac{1}{14} = \frac{2}{210} = \frac{1}{105}$