Chapter 9.7, Problem 31E

To determine

To calculate: The probability of drawing 2 red marbles from a bad that contains one green, two yellow and three red marbles without replacement.

Answer to Problem 31E

The probability to draw two red marbles is $\frac{1}{5}$ .

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, $P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}$ .

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 red 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 red color marble is drawn second time and now sample space $S_2=5$ as one is already out of the bag.

So, $n\left(E_1\right)=3$ , $n\left(E_2\right)=2$ , $n\left(S_1\right)=6$ and $n\left(S_2\right)=5$ .

Probability that both red marbles are drawn is,

  $\begin{array}{c}P\left(E\right)=\frac{n\left(E_1\right)}{n\left(S_1\right)}\cdot \frac{n\left(E_2\right)}{n\left(S_2\right)}\\ =\frac{3}{6}\cdot \frac{2}{5}\\ =\frac{1}{5}\end{array}$

Thus, the probability to draw two red marbles is $\frac{1}{5}$ .