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.

You are trying to find instrument A in a laboratory. Unfortunately, someone has put both instruments A and another kind (which we shall call B) away in identical unmarked boxes mixed at random on a shelf. You know that the laboratory has 3 A’s and 7 B’s. If you take down one box, what is the probability that you get an A? If it is a B and you put it on the table and take down another box, what is the probability that you get an A this time?

The outcomes of a particular experiment that are possible to occur comes under sample space and their probability of occurrence is 1. It is also known as possibility space or sample description space.

There is total 3 A and 7 B boxes, this implies that there is total 10 boxes with each being mutually exclusive from which A box is to be selected.

The sample space for the problem is expressed as follows,

$\left(A_1 ,A_2 ,A_3 ,B_1 ,B_2 ,B_3 ,B_4 ,B_5 ,B_6 ,B_7\right)$

Each point of the obtained sample space has an equal probability of $\frac{1}{10}$.

Find the probability that the box selected is A by adding the probabilities of each possible outcomes.

$\begin{array}{rcl}p& =& \frac{1}{10}+\frac{1}{10}+\frac{1}{10}\\ & =& \frac{3}{10}\\ & & \end{array}$

Thus, the box selected is A by adding the probabilities of each possible outcomes is $\frac{3}{10}$

When the first box is selected and it is B, it is kept aside. For the second selection, there are total 3 A and 6 B boxes, this implies that there is total 9 boxes with each being mutually exclusive from which A box is to be selected.

Find the probability that the box selected is A 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 box selected is A by adding the probabilities of each possible outcomes is$\frac{1}{3}$