# Simple Probability

The probability of an event E happening, P(E), is given by the number of ways the event can occur, n(E), compared with the total number of outcomes possible n(S) (the size of the sample space).

$P(E)=\cfrac { n(E) }{ n(S) }$

If P(E) = 0 the event is impossible, and if P(E) = 1 the event is certain.

$0 \le P(E) \le 1$

The sum of all (mutually exclusive) probabilities is 1.

### Examples

1. A container holds 8 blue, 7 white and 10 yellow marbles. If one marble is selected at random, find the probability of getting

a) a white marble

b) a white or blue marble

c) a yellow, white or blue marble

d) a red marble

The size of the sample space, or total number of marbles is 8 + 7 + 10 = 25.

a) $P(W)=\cfrac{7}{15}$

b) $P(W or B)=\cfrac{8+7}{25}=\cfrac{15}{25}=\cfrac{3}{5}$

c) $P(W or B)=\cfrac{8+7+10}{25}=\cfrac{25}{25}=1$

d) $\cfrac{0}{15}$ chance - impossible.

2. The probability that a traffic light will turn green as a car approaches it is estimated to be $\cfrac{5}{12}$. A taxi goes through 288 intersections where there are traffic lights. How many of these would be expected to turn green as the taxi approached?

As the expected chance is only $\cfrac{5}{12}$, then the theoretical solution is:

$\cfrac{5}{12} \times 288 = 120$ times.

## Complementary Events

The complement of an event happening is the event not happening. That is , the complement of P(E) is P(not E). We can write this as $P\widetilde { (E) }$.

For example:

A die is thrown. Find the probability of:

a) throwing a 4.

b) not throwing a 4.

a) $P(4)=\cfrac{1}{6}$

b) $P(not 4)=P(1, 2, 3, 5, 6)=\cfrac{5}{6}$

In general,

$P(E)=1-P\widetilde { (E) }$

## Non-Mutually Exclusive Events

Sometimes, there is an overlap where more than one event can occur at the same time. We call these non-mutually exclusive events. It is important to count the possible outcomes carefully when this happens - and Venn diagrams help.

For example:

One card is drawn from a set of cards numbered 1 to 12. Find the probability of drawing out an odd number OR a multiple of 3.

The odd cards are 1, 3, 5, 7, 9, 11.

The multiples of 3 are 3, 6, 9 and 12.

The numbers 3 and 9 are both odd and multiples of 3 (- try not to count them twice).

So there are 8 numbers that are odd OR multiples of 3: 1, 3, 5, 6, 7, 9, 11, 12.

P(odd or multiples of 3) = $\cfrac{6}{10}=\cfrac{3}{5}$

We can also display this in a Venn Diagram:

In fact, there is a formula that can be used for non-mutually exclusive events.

$P(A or B)=P(A)+P(B)-P(A and B)$

From 50 cards numbered from 1 to 50, one is selected at random. Find the probability that the card selected is even or less than 15.

Some cards are both even and less than 15 = 2, 4, 6, 8, 10, 12, 14

P(even and < 15) = $\cfrac{7}{50}$

P(even) = $\cfrac{25}{50}$

P(< 15) = $\cfrac{14}{50}$

P(even or < 15) = P(even) + P(< 15) - P(even and < 15)

P(even or < 15) = $\cfrac{25}{50} + \cfrac{14}{50} - \cfrac{7}{50}$

P(even or < 15) = $\cfrac{32}{50} = \cfrac{16}{25}$