Basic Probability

BASIC PROBABILITY Stats 102 A. Robertson Throughout the lecture below, let E and F be two sets of outcomes of an experiment. We will develop 3 rules that will aid in solving basic probability problems. These rules are: • The NOT Rule: Prob(E) = 1− Prob(NOT E) • The Multiplication Rule: Prob(E and F ) = Prob(E) × Prob(F |E). • The Addition Rule: Prob(E or F ) = Prob(E) + Prob(F ) − Prob(E and F ). THE NOT RULE Consider the simple statement: Either E happens or E does not happen. This statement is always true, regardless of what E is. Hence, Prob(E happens) + Prob(E does NOT happen) = 1. Rearranging this equation gives us the NOT Rule: Prob(E) = 1 − Prob(NOT E). This is a very simple statement, but can make all the difference when it comes to calculating some probabilities since sometimes it is easier to calculate what you don’t want to happen than it is to calculate what you do want to happen.

CALCULATING BASIC PROBABILITY We will typically assume that all possible outcomes of an experiment are equally likely. That is, no particular outcome is more likely to occur than others. In this situation, we calculate the probability of E by number of outcomes in E . Prob(E) = number of possible outcomes of the experiment We are focusing on this probability since the basic situations of statistics involve sampling and surveying. These are done by picking samples at random, which means that all samples have an equally likely chance of being picked.

CONDITIONAL PROBABILITY Probabilities can (not must) change if additional information is known. For example, the probability that Colgate will win its next football game will obviously change if we find out that every player on Colgate’s football team is sick with the flu. The probabilities we are interested here are called conditional probabilities. These are probabilities given some information. In math, these look like: Prob(E|F ) and we say “the probability of E given F.” What this is asking for is the probability that E occurs if we know that F has occurred.

Let’s use what we know to find a formula for this probability. Since we are given that E has occurred, the possible outcomes are the number of outcomes in E. And we want: of those outcomes in E, how many of them are in F . In other words, how many are in E and F . Thus, Prob(F |E) =

number of outcomes in E AND F number of outcomes in E

by the way we calculate basic probability. Now, I can divide the numerator and denominator of this fraction by the same number and not change anything. So, we have Prob(E|F ) =

(# of outcomes in E AND F )/(total # of possible outcome of the experiment) . (# of outcomes in E)/(total # of possible outcome of the experiment)

But this is, in math, Prob(F |E) =

Prob(E and F ) . Prob(E)

MULTIPLICATION RULE We are now ready to present our second rule. We use this rule to calculate the probability that both E and F occur. Rearranging the previous equation, we have Prob(E and F ) = Prob(E) × Prob(F |E). This is the multiplication rule. Before doing some examples, we consider a special case of the Multiplication Rule. If Prob(F |E) = Prob(E) (and hence, Prob(E|F ) = Prob(E)) we say that E and F are independent. In this case, the multiplication rule becomes Prob(E and F ) = Prob(E) × Prob(F ). A few examples are in order. • and flip Example 1. Roll a fair die and flip a fair coin. What is the probability that your roll a 2 • a head? Let E be the outcome that we roll a 2and let F be the outcome that we flip a head. Then 1 Prob(E and F ) = Prob(E)× Prob(F |E) = 16 × 12 = 12 .

Example 2. Shuffle a deck of 52 cards. What is the probability that the first two cards are aces? Let E be the outcome that the first card is an ace, and F the outcome that the second card is an 4 3 ace. Then Prob(E and F ) = Prob(E)× Prob(F |E) = 52 × 51 . Remark. The outcomes in Example 1 are independent since Prob(F |E) =Prob(F ), but the out3 4 while Prob(F ) = 52 . comes in Example 2 are not independent since Prob(F |E) = 51 THE ADDITION RULE Another common probability is in the case when we want either E or F (possibly both) to occur. I.e., we want Prob(E or F ). 2

To do this, we use a Venn Diagram. The rectangle encloses all possible outcomes of the experiment and the circles labeled E and F enclose all outcomes in E and F , respectively. What we want to count are all the outcomes in E or F . This is in the shaded area below.

Now, there are three regions to count. These are broken down below:

We start by counting the outcomes in E plus the outcomes in F and then writing each in terms of the three desired regions.

Substracting

from each side we get:

Translating this into math, we get the addition rule: Prob(E or F ) = Prob(E) + Prob(F ) − Prob(E and F ). Before ending our lecture, we have one more topic to cover. We say that two outcomes are mutually exclusive if Prob(E and F ) = 0. In other words, E and F cannot both occur at the same time. In this case, the addition rule reduces to Prob(E or F ) = Prob(E) + Prob(F ). Remark. A very common mistake is to use the above form of the addition rule when E and F are not mutually exclusive. By doing this, the resulting answer will always be wrong.

3

SOME EXAMPLES 1. Pick three cards, without replacement, from a standard deck. What is the probability of getting at least one ace? Solution. Our outcome E is “getting at least one ace.” Hence, this consists of getting exactly one, exactly two, or exactly three aces. Since we have three seperate cases to count, let’s look and see if the outcome (not E) has fewer cases. Here, we see that (not E) is “not getting at least one ace,” which is the same as “getting no ace.” This is easier to count. From here we use the multiplication principle, with the experiments: A is “the first card is not an ace” B is “the second card is not an ace” C is “the third card is not an ace” So we want the prob. of A and B and C. Using the multiplication rule twice we have Prob(A and B and C) = Prob(A) × Prob( B given A) × Prob(C given A and B) = (48/52) × (47/51) × (46/50). But we have counted the experiment (not E), so the answer is 1 − (48/52) × (47/51) × (46/50). 2. Roll two twenty sided dice; one blue and one red. Find the probability that you roll a blue 5 or that the sum of the two dice is even. Solution. Let E be the outcome “get a blue 5” and let F be the outcome “the sum is even.” We want Prob(E or F ) so we use the addition rule. First, Prob(E) = 1/20 and Prob(F ) = 1/2. Next, we need to find Prob(E and F ). This means we need the blue die to show 5 and the sum of the two dice to be even. Using the multiplication rule we have Prob(E and F ) = Prob(E) × Prob(F given E) = (1/20) × (1/2), where the 1/2 comes from the fact that only the numbers 1, 3, 5, 7, 9, 11, 13, 15, 17, and 19 on the red die will give an even sum when added to the blue 5. Hence, by the addition rule the answer is 1/20 + 1/2 − (1/2)(1/20) = 21/40. 3. A drawer contains 7 different pairs of socks, the pairs being colored red, blue, green, yellow, black, purple, and brown, with the socks all mixed up. You reach in and pick two socks at random. What is the probability that you get a matching pair? What is the probability that you get the brown pair? Solution. Let E be the outcome of your first pick, and let F be the outcome of your second pick. We first answer the question: What is the probability that you get a matching pair? Since, we just want a pair (any pair), it does not matter what E is. What matters is that F matches E. So reach in and pick a sock. We now want to reach in and get the same color sock from those remaining. Since there are 13 socks remaining with only one that matches, the probability of a match is 1/13. Next, we answer the question: What is the probability that you get the brown pair? Here, we have E being the outcome “the first sock is brown” and F being the outcome “the second sock is brown”. We want Prob(E and F ) so we use the multiplication principle. We have Prob(E) = 2/14 = 1/7 since two of the fourteen socks are brown. Next, we have Prob(F given E) = 1/13. Hence, the answer is (1/7) × (1/13) = 1/91. You could have answered this in a different way using your answer to the first question. Since the probability of getting a pair is 1/13 and we have seven pairs, then the probability of getting the brown pair given that you have a pair is 1/7. Again, by the multiplication principle we have (1/7) × (1/13) = 1/91.

4