Section 7.1. Probability of an Event We first define these key terms: An experiment is a procedure that yields one of a given set of possible outcomes.

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Agenda Lecture Content: Discrete Probability
Presentation transcript:

Section 7.1

Probability of an Event We first define these key terms: An experiment is a procedure that yields one of a given set of possible outcomes. The sample space of the experiment is the set of possible outcomes. An event is a subset of the sample space. Definition: If S is a finite sample space of equally likely outcomes, and E is an event, that is, a subset of S, then the probability of E is p(E) = |E|/|S|. For every event E, we have 0 ≤ p(E) ≤ 1. This follows directly from the definition because 0 ≤ p(E) = |E|/|S| ≤ |S|/|S| ≤ 1, since 0 ≤ |E| ≤ |S|. Pierre-Simon Laplace ( )

Applying Laplace’s Definition Example: An urn contains four blue balls and five red balls. What is the probability that a ball chosen from the urn is blue? Solution: The probability that the ball is chosen is 4 / 9 since there are nine possible outcomes, and four of these produce a blue ball. Example: What is the probability that when two dice are rolled, the sum of the numbers on the two dice is 7 ? Solution: By the product rule there are 6 2 = 36 possible outcomes. Six of these sum to 7. Hence, the probability of obtaining a 7 is 6 / 36 = 1 / 6.

Applying Laplace’s Definition Example: There are many lotteries that award prizes to people who correctly choose a set of six numbers out of the first n positive integers, where n is usually between 30 and 60. What is the probability that a person picks the correct six numbers out of 40 ? Solution: The number of ways to choose six numbers out of 40 is C( 40,6 ) = 40 !/( 34 ! 6 !) = 3,838,380. Hence, the probability of picking a winning combination is 1 / 3,838,380 ≈

Applying Laplace’s Definition Example: What is the probability that the numbers 11, 4, 17, 39, and 23 are drawn in that order from a bin with 50 balls labeled with the numbers 1,2, …, 50 if a) The ball selected is not returned to the bin. b) The ball selected is returned to the bin before the next ball is selected. Solution: Use the product rule in each case. a) Sampling without replacement: The probability is 1/254,251,200 since there are 50∙49∙48∙47∙46 = 254,251,200 ways to choose the five balls. b) Sampling with replacement: The probability is 1 / 50 5 = 1/312,500,000 since 50 5 = 312,500,000.

The Probability of Complements and Unions of Events Theorem 1 : Let E be an event in sample space S. The probability of the event = S − E, the complementary event of E, is given by Proof: Using the fact that | | = |S| − |E|,

The Probability of Complements and Unions of Events Example: A sequence of 10 bits is chosen randomly. What is the probability that at least one of these bits is 0 ? Solution: Let E be the event that at least one of the 10 bits is 0. Then is the event that all of the bits are 1 s. The size of the sample space S is Hence,

The Probability of Complements and Unions of Events Theorem 2 : Let E 1 and E 2 be events in the sample space S. Then Proof: Given the inclusion-exclusion formula from Section 2.2, | A ∪ B | = | A | + | B | − | A ∩ B |, it follows that

Example: What is the probability that a positive integer selected at random from the set of positive integers not exceeding 100 is divisible by either 2 or 5 ? Solution: Let E 1 be the event that the integer is divisible by 2 and E 2 be the event that it is divisible 5 ? Then the event that the integer is divisible by 2 or 5 is E 1 ∪ E 2 and E 1 ∩ E 2 is the event that it is divisible by 2 and 5. It follows that: p(E 1 ∪ E 2 ) = p(E 1 ) + p(E 2 ) – p(E 1 ∩ E 2 ) = 50/ /100 − 10/100 = 3/5. The Probability of Complements and Unions of Events

Monty Hall Puzzle Example: You are asked to select one of the three doors to open. There is a large prize behind one of the doors and if you select that door, you win the prize. After you select a door, the game show host opens one of the other doors (which he knows is not the winning door). The prize is not behind the door and he gives you the opportunity to switch your selection. Should you switch? Solution: You should switch. The probability that your initial pick is correct is 1/3. This is the same whether or not you switch doors. But since the game show host always opens a door that does not have the prize, if you switch the probability of winning will be 2/3, because you win if your initial pick was not the correct door and the probability your initial pick was wrong is 2/3. 132

The Birthday Problem What is the probability that, in a set of randomly chosen people, some pair of them will have the same birthday. By the pigeonhole principle, the probability reaches 100% when the number of people reaches 367 (since there are only 366 possible birthdays, including February 29). However, 99.9% probability is reached with just 70 people, and 50% probability with 23 people. Let’s use the compliment to explain… Consider a simulation…simulation