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1 Probability MAT230 Discrete Mathematics Fall 2018 MAT230 (Discrete Math) Probability Fall / 37

2 Outline 1 Discrete Probability 2 Sum and Product Rules for Probability 3 Expected Value MAT230 (Discrete Math) Probability Fall / 37

3 Introduction to Probability Example Suppose we roll a pair of dice and record the sum of the face-up numbers. what is the likelihood the sum is 8? This is a question about probability. Before we begin our study, we first need to define some terms. MAT230 (Discrete Math) Probability Fall / 37

4 Definitions The probability of an event is a number which expresses the long-run likelihood that the event will occur. An experiment is an activity with an observable outcome. Each repetition of an experiment is called a trial. The result of an experiment is called an outcome. The set of all possible outcomes is called the sample space. For example: Rolling a pair of dice is an experiment. We can represent the outcome with an ordered pair, such as (2, 3), to indicate the number shown on each die. We use an ordered pair rather than a set because we need to count (2, 3) and (3, 2) as two separate outcomes. This means the dice are distinguishable from one another. MAT230 (Discrete Math) Probability Fall / 37

5 The Sample Space When a pair of dice is rolled, the sample space S is given by S = {1, 2, 3, 4, 5, 6} {1, 2, 3, 4, 5, 6} so that S = 36. We can display S in a table: Die 1 Die (1,1) (1,2) (1,3) (1,4) (1,5) (1,6) 2 (2,1) (2,2) (2,3) (2,4) (2,5) (2,6) 3 (3,1) (3,2) (3,3) (3,4) (3,5) (3,6) 4 (4,1) (4,2) (4,3) (4,4) (4,5) (4,6) 5 (5,1) (5,2) (5,3) (5,4) (5,5) (5,6) 6 (6,1) (6,2) (6,3) (6,4) (6,5) (6,6) MAT230 (Discrete Math) Probability Fall / 37

6 The Sample Space In our original example we wanted to find the likelihood the sum of face-up numbers is 8 when a pair of dice are rolled. If E is the event the sum of face-up numbers on the two dice is 8, we can count the ordered pairs that sum to 8 to find E. Die (1,1) (1,2) (1,3) (1,4) (1,5) (1,6) 2 (2,1) (2,2) (2,3) (2,4) (2,5) (2,6) 3 (3,1) (3,2) (3,3) (3,4) (3,5) (3,6) 4 (4,1) (4,2) (4,3) (4,4) (4,5) (4,6) 5 (5,1) (5,2) (5,3) (5,4) (5,5) (5,6) 6 (6,1) (6,2) (6,3) (6,4) (6,5) (6,6) There are five ordered pairs that sum to 8: (2,6), (3,5), (4,4), (5,3), and (6,2) so E = 5. MAT230 (Discrete Math) Probability Fall / 37

7 Probability of Equally Likely Outcomes Definition Given an experiment with a sample space S of equally likely outcomes and an event E, the probability of the event is computed P(E) = E S. In our dice-rolling example we note that each outcome in S is equally likely (assuming the dice are not loaded) so we can compute P(E) = 5 36 = This means that when many trials are conducted, we would expect the sum of face-up numbers will be 8 just under 14% of the time. MAT230 (Discrete Math) Probability Fall / 37

8 Example 2 Example Roll a pair of dice. What is the probability that they show the same number? MAT230 (Discrete Math) Probability Fall / 37

9 Example 2 Example Roll a pair of dice. What is the probability that they show the same number? The sample space is the same as before so S = 36. The event E is both dice show the same number. We can proceed two different ways: 1 Count the ordered pairs in S of the form (k, k). There are six of them: (1, 1), (2, 2),..., (6, 6) so E = 6. 2 Use combinatorics: E = (6 possibilities on first die) (1 way to match the first die) = 6. Either way, we can compute P(E) = 6/36 = 1/6. If the experiment was repeated many times, we d expect a roll producing a pair would occur 1/6 16.7% of the time. MAT230 (Discrete Math) Probability Fall / 37

10 Example 3 Example What is the probability that when a fair coin is tossed ten times it comes up heads exactly 5 times? MAT230 (Discrete Math) Probability Fall / 37

11 Example 3 Example What is the probability that when a fair coin is tossed ten times it comes up heads exactly 5 times? The sample space S is the set of all outcomes of ten tosses and E is the subset of S corresponding to there being 5 heads and 5 tails in the ten tosses. Elements in S could be represented as strings consisting of H and T. For example HTTHHTTHTH is an element of both S and E. S = 2 10 = 1024, E = C(10, 5) = 252 P(E) = We therefore expect exactly five heads nearly 25% of the time. MAT230 (Discrete Math) Probability Fall / 37

12 Example 4 Example A bin containing 24 apples has 6 apples with worms, the remainder are worm-free. If 5 apples are selected without examination, what is the probability that 1 all are worm-free? 2 at most one has a worm? 3 all have worms? MAT230 (Discrete Math) Probability Fall / 37

13 Example 4 Example A bin containing 24 apples has 6 apples with worms, the remainder are worm-free. If 5 apples are selected without examination, what is the probability that 1 all are worm-free? 2 at most one has a worm? 3 all have worms? Solution: Let S be the set of all outcomes when 5 apples are chosen from a group of 24 so S = C(24, 5) = 42, Let E be the event all chosen apples are worm-free. In this case we re only choosing from the 24 6 = 18 worm-free apples so E = C(18, 5). P(E) = C(18, 5) 8, 568 = C(24, 5) 42, 504 = MAT230 (Discrete Math) Probability Fall / 37

14 Example 4 (Continued) Example (Continued) 2 Let E be at most one of the 5 chosen apples has a worm. Then E = (# ways to choose only worm-free apples) + (# ways to choose 1 wormy apple and 4 worm-free apples) = C(18, 5) + C(6, 1) C(18, 4) = 8, , 360 = 26, , 928 P(E) = 42, 504 = Let E be all 5 apples have worms. Then E = C(6, 5) = 6 6 P(E) = 42, 504 = 1 7, MAT230 (Discrete Math) Probability Fall / 37

15 Complementary Events Suppose S is the sample space of an experiment and E is the set of outcomes comprising a certain event. We say the complement of E is the event that E does not occur and note that E = S E. Then P(E) = E S = S E S = S E S = 1 E S = 1 P(E). Given an event E, the event E either occurs or it does not occur; one or the other must happen so P(E) + P(E) = 1. Sometimes it may be easier to compute the probability of an event by first computing the probability the event does not occur and then subtract this value from 1. MAT230 (Discrete Math) Probability Fall / 37

16 Example 5 Example Suppose five people each pick a single digit number from {0, 1,..., 9}. What is the probability that 1 exactly two people pick the same number? 2 at least two people pick the same number? MAT230 (Discrete Math) Probability Fall / 37

17 Example 5 Example Suppose five people each pick a single digit number from {0, 1,..., 9}. What is the probability that 1 exactly two people pick the same number? 2 at least two people pick the same number? We first note that an outcome for this experiment can be represented as a string of 5 digits, so S consists of digit strings of length 5 and S = (Continued) MAT230 (Discrete Math) Probability Fall / 37

18 Example 5 (Continued) Example (Continued) 1 Let E be the event exactly two people picked the same number. Then E = (# ways to choose two locations for repeated digit) (# ways to choose four distinct digits) = C(5, 2) P(10, 4) = 50, 400. The probability that exactly two people pick the same number is given by 50, 400 P(E) = 10 5 = = (Continued) MAT230 (Discrete Math) Probability Fall / 37

19 Example 5 (Continued) Example (Continued) 2 Next, let E be the event at least two people pick the same number. In this case E would be no two people picked the same number and E = P(10, 5) = 30, 240. Then P(E) = 1 P(E) 30, 240 = = = is the probability that at least two people pick the same number. MAT230 (Discrete Math) Probability Fall / 37

20 Example 5 (Continued) Example (Continued) It s worth pointing out just how much simplier it is to use the complement in our calculuation. If E is at least two people pick the same number, So E = C(5, 2) P(10, 4) exactly 2 people pick same + C(5, 2) C(3, 2) P(10, 3)/2 exactly 2 pair pick same + C(5, 3) P(10, 3) exactly 3 people pick same + C(5, 3) P(10, 2) 2 people and 3 people pick same + C(5, 4) P(10, 2) exactly 4 people pick same + C(5, 5) P(10, 1) all 5 people pick same E = 50, , , = 69, , 760 P(E) = 10 5 = MAT230 (Discrete Math) Probability Fall / 37

21 Disjoint Events Definition Two events are disjoint if they cannot occur simultaneously. Example The result of tossing a coin cannot be both heads and tails, so E 1 = comes up heads and E 2 = comes up tails are disjoint. MAT230 (Discrete Math) Probability Fall / 37

22 Disjoint Events Example Toss a pair of dice. Suppose 1 E 1 is first die shows an even number, 2 E 2 is sum of numbers shown is 4, and 3 E 3 is both dice show odd numbers. Which of the following pairs of events are disjoint? E 1 and E 2 : E 1 and E 3 : E 2 and E 3 : MAT230 (Discrete Math) Probability Fall / 37

23 Disjoint Events Example Toss a pair of dice. Suppose 1 E 1 is first die shows an even number, 2 E 2 is sum of numbers shown is 4, and 3 E 3 is both dice show odd numbers. Which of the following pairs of events are disjoint? E 1 and E 2 : not disjoint E 1 and E 3 : E 2 and E 3 : MAT230 (Discrete Math) Probability Fall / 37

24 Disjoint Events Example Toss a pair of dice. Suppose 1 E 1 is first die shows an even number, 2 E 2 is sum of numbers shown is 4, and 3 E 3 is both dice show odd numbers. Which of the following pairs of events are disjoint? E 1 and E 2 : not disjoint E 1 and E 3 : disjoint E 2 and E 3 : MAT230 (Discrete Math) Probability Fall / 37

25 Disjoint Events Example Toss a pair of dice. Suppose 1 E 1 is first die shows an even number, 2 E 2 is sum of numbers shown is 4, and 3 E 3 is both dice show odd numbers. Which of the following pairs of events are disjoint? E 1 and E 2 : not disjoint E 1 and E 3 : disjoint E 2 and E 3 : not disjoint MAT230 (Discrete Math) Probability Fall / 37

26 Sum Rule Theorem (Sum Rule) If E 1 and E 2 are disjoint events in an experiment, the probability of E 1 or E 2 is P(E 1 or E 2 ) = P(E 1 ) + P(E 2 ) Note that P(E 1 or E 2 ) could also be written as P(E 1 E 2 ). Example From our last example: P(first die shows an even number or both dice show odd numbers) = = = 3 4 = MAT230 (Discrete Math) Probability Fall / 37

27 General Sum Rule What is the probability that a card selected at random from a deck of 52 cards is a spade or an ace? MAT230 (Discrete Math) Probability Fall / 37

28 General Sum Rule What is the probability that a card selected at random from a deck of 52 cards is a spade or an ace? Let E 1 be card is spade and E 2 be card is an ace. Then P(E 1 ) = C(13, 1) C(52, 1) = = 1 4, P(E 2) = C(4, 1) C(52, 1) = 4 52 = We cannot merely add P(E 1 ) and P(E 2 ) since E 1 and E 2 are not disjoint the card could be the ace of spades. How can we proceed? MAT230 (Discrete Math) Probability Fall / 37

29 General Sum Rule Given an experiment with a sample space S and two events E 1 and E 2, both with equally likely outcomes, This generalizes as P(E 1 or E 2 ) = E 1 E 2 S Theorem (General Sum Rule) = E 1 + E 2 E 1 E 2 S = E 1 S + E 2 S E 1 E 2 S = P(E 1 ) + P(E 2 ) P(E 1 and E 2 ). If E 1 and E 2 are any events in an experiment, the probability of E 1 or E 2 is P(E 1 or E 2 ) = P(E 1 ) + P(E 2 ) P(E 1 and E 2 ), or P(E 1 E 2 ) = P(E 1 ) + P(E 2 ) P(E 1 E 2 ). MAT230 (Discrete Math) Probability Fall / 37

30 General Sum Rule Returning to our problem: What is the probability that a card selected at random from a deck of 52 cards is a spade or an ace? Let E 1 be card is spade and E 2 be card is an ace. Then P(E 1 ) = C(13, 1) C(52, 1) = = 1 4, P(E 2) = C(4, 1) C(52, 1) = 4 52 = 1 13, so P(E 1 E 2 ) = C(1, 1) C(52, 1) = P(E 1 or E 2 ) = P(E 1 ) + P(E 2 ) P(E 1 and E 2 ) = = MAT230 (Discrete Math) Probability Fall / 37

31 Independence Definition Two events are independent if the occurrence of one event is not influenced by the occurrence or non-occurrence of the other event. Determine if E 1 and E 2 are independent events in each of the following two scenarios: A single die is tossed twice. Let E 1 be first toss is a 5 and E 2 be second toss is even. Two cards are chosen from a 52-card deck. Let E 1 be at least one card is an ace and E 2 be at least one card is a king. MAT230 (Discrete Math) Probability Fall / 37

32 Independence Definition Two events are independent if the occurrence of one event is not influenced by the occurrence or non-occurrence of the other event. Determine if E 1 and E 2 are independent events in each of the following two scenarios: A single die is tossed twice. Let E 1 be first toss is a 5 and E 2 be second toss is even. Yes, these are independent events. The outcome of E 1 has no impact on E 2 or vice-versa (assuming the die is fair). Two cards are chosen from a 52-card deck. Let E 1 be at least one card is an ace and E 2 be at least one card is a king. MAT230 (Discrete Math) Probability Fall / 37

33 Independence Definition Two events are independent if the occurrence of one event is not influenced by the occurrence or non-occurrence of the other event. Determine if E 1 and E 2 are independent events in each of the following two scenarios: A single die is tossed twice. Let E 1 be first toss is a 5 and E 2 be second toss is even. Yes, these are independent events. The outcome of E 1 has no impact on E 2 or vice-versa (assuming the die is fair). Two cards are chosen from a 52-card deck. Let E 1 be at least one card is an ace and E 2 be at least one card is a king. No. If we know that one card is an ace, and therefore not a king, it reduces the probability that one of the cards is a king. MAT230 (Discrete Math) Probability Fall / 37

34 Product Rule Theorem If E 1 and E 2 are independent events in a given experiment, the probability that both E 1 and E 2 occur is P(E 1 and E 2 ) = P(E 1 ) P(E 2 ). Example A single die is tossed twice. Let E 1 be first toss is a 5 and E 2 be second toss is even. The probability that both E 1 and E 2 occur is P(E 1 and E 2 ) = = How could we proceed if the E 1 and E 2 are dependent? MAT230 (Discrete Math) Probability Fall / 37

35 Conditional Probability and the General Product Rule Definition (Conditional Probability) Given events E 1 and E 2 for some experiment, the conditional probability of E 1 given E 2, denoted P(E 2 E 1 ), is the probability that E 2 occurs given that E 1 occurs. Theorem (General Product Rule) If E 1 and E 2 are any events in a given experiment, then P(E 1 and E 2 ) = P(E 1 ) P(E 2 E 1 ), P(E 2 E 1 ) = P(E 1 and E 2 ) P(E 1 ) or = P(E 1 E 2 ) P(E 1 ) MAT230 (Discrete Math) Probability Fall / 37

36 Conditional Probability and the General Product Rule Example Example: A coin is tossed three times. What is the probability that it comes up heads all three times given that it comes up heads the first time? MAT230 (Discrete Math) Probability Fall / 37

37 Conditional Probability and the General Product Rule Example Example: A coin is tossed three times. What is the probability that it comes up heads all three times given that it comes up heads the first time? Solution: E 1 = heads occurs first time, E 2 = heads occurs three times S = 2 3 = 8, E 1 = 4, E 2 = 1, E 1 E 2 = 1 P(E 1 E 2 ) = 1 8, P(E 1) = 4 8 = 1 2 P(E 2 E 1 ) = 1/8 1/2 = 1 4 = This answer makes sense. If the first toss is known to be an head, we just need the next two out of two tosses to be heads, and the probability of that is 1/4. MAT230 (Discrete Math) Probability Fall / 37

38 General Product Rule Example Two cards are chosen from a 52-card deck. Let E 1 be at least one card is an ace and E 2 be at least one card is a king. Note that P(E 1 ) = P(E 2 ). P(E 1 ) = P(at least one card is an ace) = 1 P(neither card is an ace) = 1 C(48, 2) C(52, 2) = = P(E 1 and E 2 ) = C(4, 1)C(4, 1) C(52, 2) = = P(E 2 E 1 ) = 8/663 33/221 = MAT230 (Discrete Math) Probability Fall / 37

39 Bernoulli Trials Some experiments have only two possible outcomes, e.g. tossing a coin; such experiments are called Bernoulli Trials. We can easily compute the probability that one of these outcomes will occur a particular number of times in a sequence of trials. Suppose an experiment has only two possible outcomes. Let p be the probability of success (the stipulated event does occur) and q be the probability of failure (the event does not occur). Notice that q = 1 p. If there are to be n trials of the experiment then the probability of k successes in the n trials is given by P B = C(n, k)p k q n k. MAT230 (Discrete Math) Probability Fall / 37

40 Bernoulli Trials Example What is the probability that when a fair coin is tossed ten times it comes up heads exactly 5 times? MAT230 (Discrete Math) Probability Fall / 37

41 Bernoulli Trials Example What is the probability that when a fair coin is tossed ten times it comes up heads exactly 5 times? Solution: In this case p = q = 0.5 so P(heads exactly 5 times in 10 tosses) = C(10, 5)(0.5) 5 (0.5) We therefore expect exactly five heads nearly 25% of the time. MAT230 (Discrete Math) Probability Fall / 37

42 Bernoulli Trials Example Suppose the probability a certain baseball player will get a hit during each at-bat is 1/3. What is the probability the player gets exactly one hit in four at-bats? MAT230 (Discrete Math) Probability Fall / 37

43 Bernoulli Trials Example Suppose the probability a certain baseball player will get a hit during each at-bat is 1/3. What is the probability the player gets exactly one hit in four at-bats? Solution: We take p = 1/3 so q = 2/3. Then P(1 hit in 3 at-bats) = C(4, 1) 8 = ( ) 1 1 ( ) MAT230 (Discrete Math) Probability Fall / 37

44 Introduction to Expected Values Consider the experiment: A fair coin is tossed five times and the outcome is recorded. If this experiment was repeated many times, what is the average number of heads that would show up in each experiment? We can think of this as a weighted average using probabilities as weights. The probabilities are P(0) = C(5,0) = 1 C(5,1) , P(1) = = 5 C(5,2) , P(2) = = , P(3) = C(5,3) = C(5,4), P(4) = = 5 C(5,5) , P(5) = = Using these as weights, the average number of heads is computed = 80 = 2.5 heads 32 MAT230 (Discrete Math) Probability Fall / 37

45 Random Variables Definition Suppose the set S is the sample space of an experiment. A random variable is a function X : S R from the sample space to the real numbers. Note that a random variable is not random and is not a variable! For example, we can define a random variable X for our toss a coin five times experiment so that X returns the number of heads recorded in 5 tosses. Domain is the set of strings of length five consisting of T and H. Range is {0, 1, 2, 3, 4, 5}. Some examples: X (HTTHT) = 2, X (TTTHT) = 1, X (HHHHH) = 5. MAT230 (Discrete Math) Probability Fall / 37

46 Expected Value We use the notation P(X = x o ) to mean the probability that X is has the value x o. In our coin-tossing experiment, P(X = 2) is the probability that heads occurs exactly twice in five tosses of a fair coin. Definition For a given probability experiment, let X be a random variable whose possible values are from the set of numbers {x 1,..., x n }. Then the expected value of X, denoted E[X ], is the sum E[X ] = x 1 P(X = x 1 ) + x 2 P(X = x 2 ) + + x n P(X = x n ) n = x i P(X = x i ) i=1 MAT230 (Discrete Math) Probability Fall / 37

47 Example 1 Suppose a pair of dice is tossed. What is the expected value of the sum of numbers shown on the two dice? MAT230 (Discrete Math) Probability Fall / 37

48 Example 1 Suppose a pair of dice is tossed. What is the expected value of the sum of numbers shown on the two dice? Let X map ordered pairs of numbers shown on the dice to the sum of the numbers, i.e., X (m, n) = m + n where m, n {1, 2, 3, 4, 5, 6}. Then P(X = 2) = 1/36, P(X = 3) = 2/36, P(X = 4) = 3/36, P(X = 5) = 4/36, P(X = 6) = 5/36, P(X = 7) = 6/36, P(X = 8) = 5/36, P(X = 9) = 4/36, P(X = 10) = 3/36, P(X = 11) = 2/36, P(X = 12) = 1/36. MAT230 (Discrete Math) Probability Fall / 37

49 Example 1 Suppose a pair of dice is tossed. What is the expected value of the sum of numbers shown on the two dice? Let X map ordered pairs of numbers shown on the dice to the sum of the numbers, i.e., X (m, n) = m + n where m, n {1, 2, 3, 4, 5, 6}. Then P(X = 2) = 1/36, P(X = 3) = 2/36, P(X = 4) = 3/36, P(X = 5) = 4/36, P(X = 6) = 5/36, P(X = 7) = 6/36, P(X = 8) = 5/36, P(X = 9) = 4/36, P(X = 10) = 3/36, P(X = 11) = 2/36, P(X = 12) = 1/36. The expected value of the sum is 1 E[X ] = = 252 = 7 (average sum over many repetitions). 36 MAT230 (Discrete Math) Probability Fall / 37

50 Example 2 The Massachusetts State Lottery has a game called Mass Cash that involves trying to match 5 numbers chosen at random from 35. Matching all 5 numbers will win \$100,000 Matching any 4 numbers will win \$250 Matching any 3 numbers will win \$10 It costs \$1 to play the game. 1 What is the expected value of your winnings? 2 How much should you expect to earn? The first question is asking: If you play this game over and over, how much money would you expect to gain or lose? MAT230 (Discrete Math) Probability Fall / 37

51 Example 2 Let X be the number of matching numbers. Then P(X = 5) = P(X = 4) = P(X = 3) = C(5, 5) C(35, 5) = 1 324, 632 C(5, 4) C(30, 1) = 150 C(35, 5) 324, 632 C(5, 3) C(30, 2) = 4350 C(35, 5) 324, and E[X ] = 100, 000 P(X = 5) P(X = 4) + 10 P(X = 3) This means that our expected winnings are about \$0.56 each time we play the game. However, since we pay \$1 to play, we expect to lose about \$0.44 each time we play. MAT230 (Discrete Math) Probability Fall / 37

52 Expected Value in Trials of Independent Events Theorem Suppose an experiment has n independent trials each with probability of success p. If X is the number of successful trials then E[X ] = np In particular, this means that computing the expected value of success in a Bernoulli trial experiment requires only the product np. Example Suppose the probability a family has a baby girl is What is the expected number of girls in a family with five children? MAT230 (Discrete Math) Probability Fall / 37

53 Expected Value in Trials of Independent Events Theorem Suppose an experiment has n independent trials each with probability of success p. If X is the number of successful trials then E[X ] = np In particular, this means that computing the expected value of success in a Bernoulli trial experiment requires only the product np. Example Suppose the probability a family has a baby girl is What is the expected number of girls in a family with five children? Expected number of girls is = MAT230 (Discrete Math) Probability Fall / 37

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