Set up an appropriate sample space for each of Problems 1.1 to 1.10 and use itto solve the problem. Use either a uniform or non-uniform sample space or try both.

In a box there are 2 white, 3 black, and 4 red balls. If a ball is drawn at random,what is the probability that it is black? That it is not red?

The given data is listed as below:

• Number of white balls is, 2
• Number of black balls is, 3
• Number of red balls is, 4

The chances of occurrence of an event is equal, then the sample space is said to be the uniform sample space and if the chances of occurrence of an event is not equal then the sample space is said to be the non-uniform sample space

The total number of outcomes possible is same as the total number of balls present in the container, that is 9.

Let W represents white color balls, B represents black color balls, and R represents red color balls. So, the sample space for the problem is expressed as follows,

$\left(W_1 , W_2 , B_1 , B_2 , B_3 , R_1 , R_2 , R_3 , R_4\right)$

Each point of the obtained sample space has an equal probability of $\frac{1}{9}$.

There are 3 black balls, and each have a probability of $\frac{1}{9}$.

Find the probability that the ball selected is black by adding the probabilities of each possible outcomes.

$\begin{array}{rcl}p& =& \frac{1}{9}+\frac{1}{9}+\frac{1}{9}\\ & =& \frac{3}{9}\\ & =& \frac{1}{3}\\ & & \end{array}$

Thus, the probability that the ball selected is black is$\frac{1}{3}$.

The ball selected should not be red implies that the ball must be either white or black, so, the number of favourable outcomes is sum of number of black and white ball.

There are 3 black balls and 2 white balls, and each have a probability of $\frac{1}{9}$.

Find the probability that the ball selected is black by adding the probabilities of each possible outcomes.

$\begin{array}{rcl}p& =& \frac{1}{9}+\frac{1}{9}+\frac{1}{9}+\frac{1}{9}+\frac{1}{9}\\ & =& \frac{5}{9}\\ & & \end{array}$

Thus, the probability that the ball selected is black is$\frac{5}{9}$.