**Probability** is a measure of the likelihood that an event will occur. The probability of an event is always a number between 0 and 1 inclusive. If the event will not happen, the probability is 0. The larger the probability, the more likely it is that the event will occur, and if the event will certainly happen, the probability is 1.

We will start by looking at some simple examples of probability, where our events are equally likely to occur, and exactly one of them will occur at a time. This happens when we roll a single die, flip a single coin, or choose a ball at random from a bag.

Suppose we have a set $S$ of $n$ equally likely, mutually exclusive events, and we know that exactly one of them will occur. Let $p$ be the probability that a particular event occurs. Then since the probability of one of the events occurring is 1, we have $n \times p = 1$, so $p = \frac{1}{n}$.

If $X \subseteq S$ and we want to know what the probability that an event in $X$ occurs, we use the formula

Probability of equally likely outcomes: $p(X) = \frac{\mbox{desired events}}{\mbox{total events}} = \frac{|X|}{|S|}$

## 1. If we roll two six-sided dice, the smallest sum we can get is 2 (rolling a 1 on each die), and the largest sum we can get is 12 (rolling a 6 on each die). What is the probability of rolling a sum of 2,3, or 4?

If we look at the ordered pairs of numbers that occur, then by the Rule of Sum, there are $6\times 6 = 36$ possible outcomes. Of these outcomes, one will give a sum of 2 (the pair $(1,1)$), two will give a sum of 3 ((2,1) and (1,2)), and three will give a sum of 4 ((3,1), (2,2) and (1,3)).

Therefore, the probability of rolling a sum of $2,3,$ or $4$ is $\frac{1}{36} + \frac{2}{36} + \frac{3}{36} = \frac{1}{6}$.

## 2. In the board game monopoly, you may end up “in jail”. To get out of jail, you roll two six-sided dice, and need to get the same number on both of them. What is the probability that on a particular roll you will get out of jail?

Solution: We use the formula $p(X) = \frac{\mbox{desired outcomes}}{\mbox{total outcomes}}$ to calculate the probability. There are 36 possible ordered pairs that can occur when we roll two six-sided dice. Six of those pairs have the same number on both dice $( (1,1), (2,2), \ldots, (6,6) )$, so the probability is $\frac{6}{36} = \frac{1}{6}$.

## 3. A bag contains 4 red balls, 5 blue balls, and 3 black balls. If three balls are drawn at random from the bag, what is the probability all three balls are different colours?

Solution: There are $4 + 5 + 3 = 12$ balls in the bag, so there are $\binom{12}{3} = 220$ ways to choose three balls from the bag. If we want one ball of each colour, by the rule of product there are $4 \times 5 \times 3 = 60$ ways we can choose them. So the probability is $\frac{60}{220} = \frac{3}{11}$.

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