## Introduction to Probability

First, we will discuss experiments where outcomes are equally likely to occur and the frequency approach to assigning probabilities. Then, we will focus on the concept of events and touch on the issue of conditional probability.

##### Learning Objectives

1. Compute probability in a situation where there are equally-likely outcomes
2. Apply concepts to cards and dice
3. Compute the probability of two independent events both occurring
4. Compute the probability of either of two independent events occurring
5. Do problems that involve conditional probabilities
6. Compute the probability that in a room of N people, at least two share a birthday
7. Describe the gambler's fallacy
8. Probability of a Single Event

If you roll a six-sided die, there are six possible outcomes, and each of these outcomes is equally likely. A six is as likely to come up as a three, and likewise for the other four sides of the die. What, then, is the probability that a one will come up? Since there are six possible outcomes, the probability is 1/6. What is the probability that either a one or a six will come up? The two outcomes about which we are concerned (a one or a six coming up) are called favorable outcomes. Given that all outcomes are equally likely, we can compute the probability of a one or a six using the formula:

$probability =\frac{\text { Number of favorable outcomes }}{\text { Number of possible equally-likely outcomes }}$

In this case there are two favorable outcomes and six possible outcomes. So the probability of throwing either a one or six is 1/3. Don't be misled by our use of the term "favorable," by the way. You should understand it in the sense of "favorable to the event in question happening". That event might not be favorable to your well-being. You might be betting on a three, for example.

The above formula applies to many games of chance. For example, what is the probability that a card drawn at random from a deck of playing cards will be an ace? Since the deck has four aces, there are four favorable outcomes; since the deck has 52 cards, there are 52 possible outcomes. The probability is therefore 4/52 = 1/13. What about the probability that the card will be a club? Since there are 13 clubs, the probability is 13/52 = 1/4.

Let's say you have a bag with 20 cherries: 14 sweet and 6 sour. If you pick a cherry at random, what is the probability that it will be sweet? There are 20 possible cherries that could be picked, so the number of possible outcomes is 20. Of these 20 possible outcomes, 14 are favorable (sweet), so the probability that the cherry will be sweet is 14/20 = 7/10. There is one potential complication to this example, however. It must be assumed that the probability of picking any of the cherries is the same as the probability of picking any other. This wouldn't be true if (let us imagine) the sweet cherries are smaller than the sour ones. (The sour cherries would come to hand more readily when you sampled from the bag.) Let us keep in mind, therefore, that when we assess probabilities in terms of the ratio of favorable to all potential cases, we rely heavily on the assumption of equal probability for all outcomes.

Here is a more complex example. You throw 2 dice. What is the probability that the sum of the two dice will be 6? To solve this problem, list all the possible outcomes. There are 36 of them since each die can come up one of six ways. The 36 possibilities are shown below.

 Die 1 Die 2 Total Die 1 Die 2 Total Die 1 Die 2 Total 1 1 2 3 1 4 5 1 6 1 2 3 3 2 5 5 2 7 1 3 4 3 3 6 5 3 8 1 4 5 3 4 7 5 4 9 1 5 6 3 5 8 5 5 10 1 6 7 3 6 9 5 6 11 2 1 3 4 1 5 6 1 7 2 2 4 4 2 6 6 2 8 2 3 5 4 3 7 6 3 9 2 4 6 4 4 8 6 4 10 2 5 7 4 5 9 6 5 11 2 6 8 4 6 10 6 6 12

You can see that 5 of the 36 possibilities total 6. Therefore, the probability is 5/36.

If you know the probability of an event occurring, it is easy to compute the probability that the event does not occur. If $P(A)$ is the probability of Event A, then $1 - P(A)$ is the probability that the event does not occur. For the last example, the probability that the total is 6 is 5/36. Therefore, the probability that the total is not 6 is 1 - 5/36 = 31/36.