Theoretical Possibility Spaces for the ESAT

Possibility Spaces and Theoretical Probability

Possibility spaces, often referred to as sample spaces, are systematic methods used to represent all the different possible outcomes of a probability experiment. They are particularly useful when experiments involve equally likely outcomes, as they allow for the precise calculation of theoretical probabilities.

These spaces can be constructed for single experiments or for combined experiments where two or more events occur together. For instance, consider flipping a fair coin. The possible outcomes are heads or tails. Since these outcomes are equally likely, the probability of obtaining a head is $\frac{1}{2}$ and the probability of obtaining a tail is also $\frac{1}{2}$.

When dealing with combined experiments, such as rolling two unbiased six-sided dice, a possibility space diagram is essential. This diagram lists every possible pair of outcomes from the two events.

	1	2	3	4	5	6
1	(1,1)	(1,2)	(1,3)	(1,4)	(1,5)	(1,6)
2	(2,1)	(2,2)	(2,3)	(2,4)	(2,5)	(2,6)
3	(3,1)	(3,2)	(3,3)	(3,4)	(3,5)	(3,6)
4	(4,1)	(4,2)	(4,3)	(4,4)	(4,5)	(4,6)
5	(5,1)	(5,2)	(5,3)	(5,4)	(5,5)	(5,6)
6	(6,1)	(6,2)	(6,3)	(6,4)	(6,5)	(6,6)

This diagram demonstrates that there are 36 possible outcomes when two dice are rolled. We can use this to find the probability of specific events. For example, to find the probability of both dice showing an odd number, we identify the favourable outcomes: (1,1), (1,3), (1,5), (3,1), (3,3), (3,5), (5,1), (5,3), and (5,5). Since there are 9 such outcomes, the probability is $\frac{9}{36} = \frac{1}{4}$.

Single Experiments

In a single experiment, we list all possible values to determine the theoretical probability of different results. Consider rolling a normal six-sided die. If the die is fair, there are six equally likely outcomes: 1, 2, 3, 4, 5, and 6.

To find the probability that the die lands on a three, we use the formula: $P(\text{three}) = \frac{\text{number of favourable outcomes}}{\text{total number of possible outcomes}} = \frac{1}{6}$

To find the probability of rolling an odd number, we identify the odd outcomes (1, 3, 5). There are three favourable outcomes: $P(\text{odd number}) = \frac{3}{6} = \frac{1}{2}$

To find the probability of rolling a five or more, we identify the outcomes 5 and 6. There are two favourable outcomes: $P(\text{five or more}) = \frac{2}{6} = \frac{1}{3}$

To find the probability of rolling a square number, we identify the outcomes 1 and 4. There are two favourable outcomes: $P(\text{square number}) = \frac{2}{6} = \frac{1}{3}$

Combined Experiments

When two events are combined and their results are processed (for example, by multiplying the scores), the possibility space should reflect the final results of that process. Suppose two fair six-sided dice are rolled and their scores are multiplied together. We can construct a possibility space showing the products:

	1	2	3	4	5	6
1	1	2	3	4	5	6
2	2	4	6	8	10	12
3	3	6	9	12	15	18
4	4	8	12	16	20	24
5	5	10	15	20	25	30
6	6	12	18	24	30	36

We can use this product table to calculate specific probabilities:

1. The probability of the product being odd: By counting the odd numbers in the table, we find there are 9 such products. Therefore, $P(\text{odd product}) = \frac{9}{36} = \frac{1}{4}$.

2. The probability of the product being greater than 10 but less than 20: We look for products in the range 11 to 19. These are: 12 (four times), 15 (twice), 16 (once), and 18 (twice). There are 9 favourable outcomes in total, so $P(10 < \text{product} < 20) = \frac{9}{36} = \frac{1}{4}$.