# Introduction to probability

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1 Introduction to probability Suppose an experiment has a finite set X = {x 1,x 2,...,x n } of n possible outcomes. Each time the experiment is performed exactly one on the n outcomes happens. Assign each outcome a real number between 0 and 1, called the probability of that outcome. The probability of an outcome is supposed to be proportional to its likelihood of happening. We want a probability of an outcome being near 1 to mean that that outcome is very likely to be the one that happens, and a probability near 0 to mean that that outcome almost never happens. Write p(x i ) for the probability of the outcome x i. The sum of the probabilities of all outcomes in X must be 1 because the outcomes in X are the only possible outcomes, so one of them must happen. 1

2 So far we have 0 p(x i ) 1 for each i and n i=1 p(x i ) = 1. Where do the probabilities p(x i ) come from? Sometimes they come from doing an experiment many times and tabulating the outcomes. Example. Weather forecasters save enormous tables of weather conditions. The prediction, There is a 40% chance of rain tomorrow. means that on 40% of the days (listed in the weather records) when the weather conditions were similar to what they are now, it rained the next day. If outcome x has a 40% chance of happening, then its probability is written p(x) = 0.4 so that it will be between 0 and 1. 2

3 Sometimes we expect that all possible outcomes are equally likely because there is no reason to think that some outcomes are more likely than others. In this case, if there are n possible outcomes, then each outcome x has probability p(x) = 1/n. Example. Suppose a deck of 52 cards has been shuffled well and then one card is chosen. The probability that the chosen card is the Six of Hearts is p(six of Hearts) = 1/52. This is an example of equally likely outcomes. Here is another. Example. If a coin is properly balanced and tossed well, then the two sides Heads and Tails are equally likely, so each of these outcomes has probability

4 We will often combine some outcomes and ask for the probability that at least one of them happens, but we don t care which one. A subset E of a set X of all possible outcomes is called an event. We say that E happens if the outcome of the experiment is one of the outcomes in E. The probability of an event E is the sum of the probabilities of the outcomes in it. We write p(e) = x Ep(x). For any event, 0 p(e) 1. The probability that E does not happen is 1 p(e). 4

5 Example. In a deck of cards, 13 of the 52 cards are Clubs, so the probability that a Club is drawn from a shuffled deck is = 1 4 = Four of the cards are Jacks (one from each of the four suits), so the probability that a Jack is drawn is = Later, we will compute probabilities of events like this one: Suppose two cards are drawn from a deck. What is the probability that they are in the same suit? 5

6 Events may be combined using set theory. The union E F of events E, F, happens if either one of them happens, that is, if the outcome is in either E or F. The intersection E F of events E, F happens if both happen, that is, if the outcome is in both E and F. Two events E, F are mutually exclusive if they are disjoint sets, that is, E F is empty. In other words, E, F are mutually exclusive if they cannot both happen. When E, F are mutually exclusive, p(e F) = p(e)+p(f). (Recall the definition of p(e).) Ditto for more than two events being mutually exclusive. Example. The probability that a card drawn from a deck is either a Jack, a Queen or a King is 1/13 + 1/13 + 1/13 = 3/13 because a card may be at most one of Jack, Queen, King. 6

7 Suppose E and F are two events and F can happen, that is, p(f) > 0. Define the conditional probability of E given F to be p(e F) = p(e F). p(f) Example. Find the conditional probability that a card is a Queen given that it is either a Jack, a Queen or a King. Here, E is the event, the card is a Queen and F is the event, the card is either a Jack, a Queen or a King. We have p(e) = 1/13, p(f) = 3/13 and p(e F) = 1/13 because E F = E. The answer is p(e F) = p(e F) p(f) = 1/13 3/13 =

8 Bayes Theorem Write the definition of conditional probability on the form p(e F) = p(e F)p(F). If we interchange E and F, we get p(f E) = p(f E)p(E). Since F E = E F, we have p(e F)p(F) = p(f E)p(E), and we have proved Bayes Theorem: Theorem. If both p(e) > 0 and p(f) > 0, then p(f)p(e F) p(f E) =. p(e) 8

9 Example of Bayes Theorem. Draw one card from a deck. Let F be the event, the card is a Jack or Queen or King of Spades. Let E be the event, the card is a Queen. Since the Jack, Queen and King of Spades are 3 of the 52 cards, p(f) = 3/52. Since 4 of the 52 cards are Queens, p(e) = 4/52 = 1/13. Let us compute p(e F). If F happens, then the card is one of the three cards: Jack or Queen or King of Spades. One of these is a Queen, so p(e F) = 1/3. By Bayes Theorem, p(f E) = p(f)p(e F) p(e) = (3/52)(1/3) 1/13 = 1 4. This result is easy to verify, because if the card is a Queen, then it has 1 chance in 4 of being the Queen of Spades. Thus, p(f E) = 1/4. 9

10 Independent Events Two events E and F are called independent if p(e F) = p(e). Intuitively, this says that E and F are independent if the probability that E happens does not depend on whether F happens. When both p(e) > 0 and p(f) > 0, Bayes Theorem implies that p(e F) = p(e) if and only if p(f E) = p(f). The formula p(e F) = p(e F)p(F) and the definition of independent imply that E and F are independent iff p(e F) = p(e) p(f). 10

11 The two events in the Bayes Theorem example are not independent because p(e F) = 1/52 since the card must be the Queen of Spades, while p(e) p(f) = (1/13)(3/52) 1/52. Example. Let E be the event, the card is a Spade. Let F be the event, the card is a Queen. Since there are 13 Spades, p(e) = 13/52 = 1/4. Since there are 4 Queens, p(f) = 4/52 = 1/13. The event E F says that the card is the Queen of Spades, which is 1 of 52 cards, so p(e F) = 1/52 = (1/4)(1/13) = p(e)p(f), so the events E and F are independent. 11

12 Random Variables A sample space is the set of all possible outcomes x i, each having a probability p(x i ). (In this class, we assume the number of possible outcomes is finite.) Example. Draw a card. There are 52 possible outcomes, like x i = Queen of Spades. Each has probability p(x i ) = 1/52. The sample space is the set of 52 cards. A random variable is a (real-valued) function r defined on a sample space. Example. Draw a card x i. Let r(x i ) denote the value of the card, defined as follows: If the card has a number, this number is its value. So r(six of Clubs) = 6. If the card is an Ace, then its value is 1: r(ace of Diamonds) = 1. If the card is a Jack, Queen or King, then its value is 10: r(queen of Hearts) = 10. Then r(x i ) is a random variable define on a deck of cards. 12

13 Let r 1, r 2,..., be all possible values of a random variable r defined on a sample space. (This is a finite number of values.) The probability distribution of r is the function f defined by f(r j ) = p(r(x i ) = r j ), that is, f(r j ) is the probability of the event r(x i ) = r j. Example. In the example of the random variable on the deck of cards above, f(i) = 1/13 for 1 i 9 because 4 of the 52 cards have value i in this range. However, f(10) = 4/13 since 4 cards in each suit have value 10. Two random variables r, s are independent if for any possible values r 1, s 1, they could assume, the probability that r(x) = r 1 and s(x) = s 1 equals p(r(x) = r 1 ) p(s(x) = s 1 ). Example. Draw a card. Record its value as r. Replace the card in the deck. Shuffle the deck again and draw a second card. Record its value as s. Then r and s are independent random variables with the same probability distribution. 13

14 There are several concise ways to describe the probability distribution of a random variable by giving a typical value of it. The median of the probability distribution f of a random variable r is a value r m so that the probability of r(x) > r m is as close to 0.5 as possible. (f is used to compute this probability.) The median is the middle value of r(x). Example. Suppose r has this probability distribution: r f(r) The median of r is 6 because p(r > 6) is 0.6 while p(r > 8) is 0.2 and 0.6 is closer to

15 Another (more useful) typical value is the mean or average or expected value. The mean or expected value of a random variable r with values r 1, r 2,... and probability distribution f is µ = E(r) = i r i f(r i ). Example. The mean of the value of cards in the example above is = ( )( ) = This number, 85/13 6.5, is the average value of a card. 15

16 If F is a real function of a real variable, and r is a random variable, then F(r) is another random variable. It has value F(r(x)) on outcome x. Its expected value is µ = E(F(r)) = i F(r i )f(r i ). The k-th moment of a random variable r is the expected value of F(r) = r k. The variance of a random variable r with expected value µ is Var(r) = E((r µ) 2 ) = E(r 2 ) µ 2. The last equation is a simple theorem. The square root of the variance of r is the standard deviation of r. It measures how much r(x) varies from the mean µ. 16

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