Chapter 9.7, Problem 34E

To determine

To find: the probability for the experiment of drawing two marbles at random (without replacement) such that both are of different colors, from a bag containing one green, two yellow, and three red marbles.

Answer to Problem 34E

  $\frac{11}{15}$

Explanation of Solution

Instead of finding the probability of drawing two marbles of different colors, find the probability that both marbles are of same color and then use the fact that probability of complement of an event A is given by $P\left(A^{\prime }\right)=1-P\left(A\right)$ to find the probability of drawing two marbles of different colors.

Total number of marbles $=1+2+3=6$

Now, first find the probability that both marbles are of yellow color.

Here each draw is independent, so take product of each event to find total probability.

  $\begin{array}{c}P\left(\text{both yellow}\right)=\frac{2}{6}\times \frac{1}{5}\\ =\frac{1}{15}\end{array}$

Now, find the probability that both marbles are of red color.

Here each draw is independent, so take product of each event to find total probability.

  $\begin{array}{c}P\left(\text{both red}\right)=\frac{3}{6}\times \frac{2}{5}\\ =\frac{1}{5}\end{array}$

Then, probability of drawing two marbles of same color is:

  $\begin{array}{c}P\left(\text{same color}\right)=P\left(\text{both yellow}\right)+P\left(\text{both red}\right)\\ =\frac{1}{15}+\frac{1}{5}\\ =\frac{4}{15}\end{array}$

Lastly, the probability of drawing two marbles of different color is:

  $\begin{array}{c}P\left(\text{different color}\right)=1-P\left(\text{same color}\right)\\ =1-\frac{4}{15}\\ =\frac{11}{15}\end{array}$