Chapter 9.7, Problem 33E

To determine

To calculate: The probability of drawing 2 marbles such that neither is yellow marble from a bad that contains one green, two yellow and three red marbles without replacement.

Answer to Problem 33E

The probability to draw two marbles such that neither is yellow marble is $\frac{14}{15}$ .

Explanation of Solution

Given information:

An experiment that draws two marbles from a bag that contains one green, two yellow and three red marbles without replacement.

Formula used:

The probability of an event is the ratio of number of possible outcomes of an event E denoted by $n\left(E\right)$ and the sample space of equally likely outcome denoted by $n\left(S\right)$ . Mathematically it is denoted by, $P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}$ .

The complement of an event E is denoted by $E\text{'}$ and probability that an event does not occur is

  $P\left(E\text{'}\right)=1-P\left(E\right)$

Calculation:

Consider the provided experiment that draws two marbles from a bag that contains one green, two yellow and three red marbles without replacement.

Total number of marbles are 6.

Recall that the probability of an event is the ratio of number of possible outcomes of an event E denoted by $n\left(E\right)$ and the sample space of equally likely outcome denoted by $n\left(S\right)$ . Mathematically it is denoted by, $P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}$ .

Denote the event $E_1$ as a yellow color marble is drawn and sample space $S_1=6$ .

Since, the experiment is done without replacing the marbles and two marbles are drawn.

Denote the event $E_2$ as a yellow color marble is drawn second time and now sample space $S_2=5$ as one is already out of the bag.

So, $n\left(E_1\right)=2$ , $n\left(E_2\right)=1$ , $n\left(S_1\right)=6$ and $n\left(S_2\right)=5$ .

Probability that both yellow marbles are drawn is,

  $\begin{array}{c}P\left(E\right)=\frac{n\left(E_1\right)}{n\left(S_1\right)}\cdot \frac{n\left(E_2\right)}{n\left(S_2\right)}\\ =\frac{2}{6}\cdot \frac{1}{5}\\ =\frac{1}{15}\end{array}$

Now, probability that neither marble is yellow is probability that both marbles are yellow is subtract from one since one event is complement to other.

Denote the event $E\text{'}$ as neither yellow color marble is drawn

  $\begin{array}{c}P\left(E\text{'}\right)=1-P\left(E\right)\\ =1-\frac{1}{15}\\ =\frac{14}{15}\end{array}$

Thus, the probability to draw two marbles such that neither is yellow marble is $\frac{14}{15}$ .