## 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.

Events $A$ and $B$ are independent events if the probability of Event $B$ occurring is the same whether or not Event $A$ occurs. Let's take a simple example. A fair coin is tossed two times. The probability that a head comes up on the second toss is 1/2 regardless of whether or not a head came up on the first toss. The two events are (1) first toss is a head and (2) second toss is a head. So these events are independent. Consider the two events (1) "It will rain tomorrow in Houston" and (2) "It will rain tomorrow in Galveston" (a city near Houston). These events are not independent because it is more likely that it will rain in Galveston on days it rains in Houston than on days it does not.

##### Probability of A and B

When two events are independent, the probability of both occurring is the product of the probabilities of the individual events. More formally, if events $A$ and $B$ are independent, then the probability of both $A$ and $B$ occurring is:

$P(A\;and\;B) = P(A) x P(B)$

where $P(A\;and\;B)$ is the probability of events $A$ and $B$ both occurring, P(A) is the probability of event $A$ occurring, and $P(B)$ is the probability of event $B$ occurring.

If you flip a coin twice, what is the probability that it will come up heads both times? Event $A$ is that the coin comes up heads on the first flip and Event $B$ is that the coin comes up heads on the second flip. Since both $P(A)$ and $P(B)$ equal 1/2, the probability that both events occur is

$1/2\times1/2 = 1/4$.

Let's take another example. If you flip a coin and roll a six-sided die, what is the probability that the coin comes up heads and the die comes up 1? Since the two events are independent, the probability is simply the probability of a head (which is 1/2) times the probability of the die coming up 1 (which is 1/6). Therefore, the probability of both events occurring is 1/2 x 1/6 = 1/12.

One final example: You draw a card from a deck of cards, put it back, and then draw another card. What is the probability that the first card is a heart and the second card is black? Since there are 52 cards in a deck and 13 of them are hearts, the probability that the first card is a heart is 13/52 = 1/4. Since there are 26 black cards in the deck, the probability that the second card is black is 26/52 = 1/2. The probability of both events occurring is therefore 1/4 x 1/2 = 1/8.

See the section on conditional probabilities on this page to see how to compute $P(A\;and\;B)$ when A and B are not independent.

##### Probability of $A$ or $B$

If Events A and B are independent, the probability that either Event A or Event B occurs is:

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

In this discussion, when we say "$A$ or $B$ occurs" we include three possibilities:

1. $A$ occurs and $B$ does not occur
2. $B$ occurs and $A$ does not occur
3. Both  $A$ and $B$ occur

This use of the word "or" is technically called inclusive or because it includes the case in which both $A$ and $B$ occur. If we included only the first two cases, then we would be using an exclusive or.

(Optional) We can derive the law for $P(A-or-B)$ from our law about $P(A-and-B)$. The event "$A-or-B$" can happen in any of the following ways::

1. $A$-and-$B$ happens
2. $A$-and-not-$B$ happens
3. not-$A$-and-$B$ happens.

The simple event $A$ can happen if either $A$-and-$B$ happens or $A$-and-not-$B$ happens. Similarly, the simple event $B$ happens if either $A$-and-$B$ happens or not-$A$-and-$B$ happens. $P(A)+P(B)$ is therefore $P(A$-and-$B$ $)+P(A$-and-not-$B$ $)+$ $P(A$-and-$B)+P($ not-$A$-and-$B$ $)$, whereas $P(A$-or-$B$ $)$ is $P(A$-and-$B$ $)+P(A$-and-not$B$) $+P($ not-$A$-and-$B$ $)$. We can make these two sums equal by subtracting one occurrence of $P(A$-and-$B)$ from the first. Hence, $P(A-$or-$B)=P(A)+P(B)-P(A-$and-$B$).

Now for some examples. If you flip a coin two times, what is the probability that you will get a head on the first flip or a head on the second flip (or both)? Letting Event A be a head on the first flip and Event B be a head on the second flip, then $P(A) = 1/2$, $P(B) = 1/2$, and $P(A\;and\;B) = 1/4$. Therefore,

$P(A\;or\;B) = 1/2 + 1/2 - 1/4 = 3/4$.

If you throw a six-sided die and then flip a coin, what is the probability that you will get either a 6 on the die or a head on the coin flip (or both)? Using the formula,

$P(6\;or\;head) = P(6) + P(head) - P(6\;and\;head) = (1/6) + (1/2) - (1/6)(1/2) = 7/12$

An alternate approach to computing this value is to start by computing the probability of not getting either a 6 or a head. Then subtract this value from 1 to compute the probability of getting a 6 or a head. Although this is a complicated method, it has the advantage of being applicable to problems with more than two events. Here is the calculation in the present case. The probability of not getting either a 6 or a head can be recast as the probability of

(not getting a 6) AND (not getting a head).

This follows because if you did not get a 6 and you did not get a head, then you did not get a 6 or a head. The probability of not getting a six is 1 - 1/6 = 5/6. The probability of not getting a head is 1 - 1/2 = 1/2. The probability of not getting a six and not getting a head is 5/6 x 1/2 = 5/12. This is therefore the probability of not getting a 6 or a head. The probability of getting a six or a head is therefore (once again) 1 - 5/12 = 7/12.

If you throw a die three times, what is the probability that one or more of your throws will come up with a 1? That is, what is the probability of getting a 1 on the first throw OR a 1 on the second throw OR a 1 on the third throw? The easiest way to approach this problem is to compute the probability of

NOT getting a 1 on the first throw

AND not getting a 1 on the second throw

AND not getting a 1 on the third throw.

The answer will be 1 minus this probability. The probability of not getting a 1 on any of the three throws is 5/6 x 5/6 x 5/6 = 125/216. Therefore, the probability of getting a 1 on at least one of the throws is 1 - 125/216 = 91/216.