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Applied Calculus II

Applied Calculus II. Probability Slides subject to change. Probability. An experiment is a situation involving chance or probability that leads to results called “outcomes.” An outcome is the result of a single trial of an experiment. An event is one or more outcomes of an experiment.

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Applied Calculus II

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  1. Applied Calculus II Probability Slides subject to change

  2. Probability • An experiment is a situation involving chance or probability that leads to results called “outcomes.” • An outcome is the result of a single trial of an experiment. • An event is one or more outcomes of an experiment. • Probability is the measure of how likely an event is.

  3. Probability Equation • Probability of an event: • The sample space of an experiment is the set of all possible outcomes of that experiment.

  4. Example • Experiment: A single 6-sided die is rolled. • What is the probability of each outcome?   • Outcomes:  The possible outcomes of this experiment are 1, 2, 3, 4, 5 and 6. • What is the probability of rolling an even number? 3/6 or 1/2 • This experiment illustrates the difference between an outcome and an event. A single outcome of this experiment is rolling a 1, or rolling a 2, etc. Rolling an even number (2, 4 or 6) − possible outcome of several rolls − is an event, and rolling an odd number (1, 3 or 5) is also an event. Sample Space Probability of Each Outcome

  5. Total Number of Outcomes • Find the probability that when a couple has 3 children, they will have exactly 2 boys. Assume boys and girls are equally likely. Three correspond to exactly two boys. p(2 boys in 3 births) = 3/8 = 0.375

  6. Problem • A couple is planning to have four children. • Construct a table of possible gender outcomes. • Find the probability of getting exactly two boys and two girls. • Find the probability that all four children are boys. 0.38 0.06

  7. Probability Range • The probability of an impossible event is 0. • The probability of an event that is certain to occur is 1. • For any event A, the probability of A is between 0 and 1 inclusive, 0 ≤ p(A)≤ 1. Impossible 0 Certain1 Very Unlikely Somewhat Unlikely 50-50 Probability Rather Probable Very Likely

  8. Probability • The complement of event , denoted by consists of all outcomes in which event does not occur. • Throwing a die, 1 is p(1) = 1/6, p( ) is 5/6 (throw a 2, 3, 4, 5, or 6).

  9. Odds • Expression of likelihood are called odds (such as “50 to 1”) • Actual odds in favor, actual odds against ... • Payoff odds established by the racetrack or casino. • Odds are 3 to 1 favoring Fred to win the race. In probability terms p(Fred’s winning) = ¾ = 0.75. • If the odds are 3 to 1 against Fred’s winning, p(Fred’s winning) = ¼ = 0.25.

  10. Addition Rule • A compound event is any event combining two or more simple events. • When two events, A and B, are mutually exclusive, the probability that A or B will occur is the sum of the probability of each event.  p(A or B) = p(A) + p(B) • Nicknamed the “Or – Or Rule.” • Events are mutually exclusive if p(A and B) = 0, e.g., a die is either 1, or 2, or 3, etc.. A coin is either heads or tails.

  11. Example: Mutually Exclusive • A single 6-sided fair die is rolled. • What is the probability of rolling a 2 or a 5? • Probabilities:   • p(2) = 1/6 • p(5) = 1/6 • p(2 or 5) = p (2) + p (5) • p(2 or 5) =1/6 + 1/6 = 2/6=1/3

  12. Non-Mutually Exclusive • A single card is chosen at random from a standard deck of 52 playing cards. • What is the probability of choosing a king or a club? • A chosen card may be both king and club. This requires a different equation. • The addition causes the king of clubs to be counted twice, so its probability must be subtracted.

  13. Non-Mutually Exclusive • Probabilities:   • p(king or club) = p(king) + p(club) − p(king of clubs) • p(king or club) = 4/52 + 13/52 − 1/52 = 16/52 = 0.308 • When two events are non-mutually exclusive, a different addition rule must be used. • p(A or B) = p (A) + p (B) − p (A and B)

  14. Problems • In a pet store, there are 6 puppies, 9 kittens, 4 gerbils and 7 parakeets. If a pet is chosen at random, what is the probability of choosing a puppy or a parakeet? • The probability of a New York teenager owning a skateboard is 0.37, of owning a bicycle is 0.81 and of owning both is 0.36. If a New York teenager is chosen at random, what is the probability that the teenager owns a skateboard or a bicycle? 13/26 0.82

  15. Multiplication Rule • Two events, A and B, are independent if the fact that A occurs does not affect the probability of B occurring. • When two events, A and B, are independent, the probability of both occurring is:  p(A and B) = p (A)• p (B) • Nicknamed the “And-And Rule.”

  16. Example • Experiment:  A coin is tossed and a single 6-sided die is rolled. • Find the probability of a coin landing head−side up and rolling a 3 on the die. • Probabilities:   • p(head) = 1/2 • p(roll a 3) = 1/6 • p(head and 3) = p (head) •p (roll a 3) • = 1/2 · 1/6 = 1/12

  17. Problem • Experiment:  A jar contains 3 red, 5 green, 2 blue and 6 yellow marbles. A marble is chosen at random from the jar. After replacing it, a second marble is chosen. • What is the probability of choosing a green and a yellow marble? 1/30

  18. Conditional Probability • The conditional probability of an event B in relationship to an event A is the probability that event B occurs given that event A has already occurred. • The notation for conditional probability is p(B|A) • [pronounced as ”The probability of event B given A”].

  19. Conditional Probability • p(B|A) means the probability of event B given that event A has already occurred. • To find the probability of the two dependent events, we use a modified version of the multiplication rule. • Multiplication Rule 2:  When two events, A and B, are dependent, the probability of both occurring is:  • p(A and B)  =  p(A)  •p(B|A) Note

  20. Example • A card is chosen at random from a standard deck of 52 playing cards. Without replacing it, a second card is chosen. • What is the probability that the first card chosen is a queen and the second card chosen is a jack? • p(A) = 4/52 • p(B|A) = 4/51 note this is dependent • p(A and B) = 4/52  • 4/51 = 16/2652

  21. Example • On a math test, 5 out of 20 students got an A. If three students are chosen at random without replacement, what is the probability that all three got an A on the test? • Probabilities:   • p(3 and 2 and 1) = 5/20 · 4/19 · 3|18 • = 60/6840 = 1/114

  22. Problem • Mr. Parietti needs two students to help him with a science demonstration for his class of 18 girls and 12 boys. He randomly chooses one student who comes forward. He then chooses a second student from those still seated. What is the probability that both students chosen are girls? • Probabilities:   • p(Girl 1 and Girl 2) = p(Girl 1) and p(Girl 2|Girl1) • finish 51/145

  23. Fundamental Counting Principle • How do we count outcomes without listing them? • Restaurant offers 4 choices for main dish (chicken, beef, ham, fish), 3 choices for side dishes (soup, salad, rice) How many days will we be able to order different meals before we have to start repeating our choices? • Principle 1: If k items from one list can be combined with m items from another list, there are km possibilities. • Possible outcomes = 4 x 3 = 12

  24. Problems •  A college offers 7 courses in the morning and 5 in the evening. Find the possible number of choices for the student if he wants to study one course in the morning and one in the evening. • How many different license plates can be made by a state, if each plate is to display three letters followed by three numbers? 35 17,576,000

  25. Count Outcomes • Principle 2: If you make n choices with replacement (i.e., an item or value can be selected more than once) from a group of m possibilities, the total number of possibilities is mn. • Seven-letter password: 7 choices each with 26 possibilities. = 267 = 8.03x109. • Chance of someone guessing it is 1/ 8.03x109 = 1.25x10−10.

  26. Problem • A combination lock has 10 digits in each ring, with 3 rings. How many possible combinations are there? • Another combination lock has 40 numbers, one of which must be selected on each of three turns of the dial. Theoretically, how many possible combinations are there? • Find the number of possible outcomes out of 4 births, boys and girls being equally probable. 1,000 64,000 16

  27. Ways of Putting Objects in Order • If you have n distinct objects, then n! gives the number of ways of putting the objects in different order, where order does matter. In this case, you have to reduce the number of available choices each time. • The factorial function (symbol: !) just means to multiply a series of descending natural numbers. • 0! = 1 • 1! = 1 • 2! = 2x1= 2 • 3! = 3x2x1= 6

  28. Example • What order could 16 pool balls be in? After choosing, say, number "14" you can't choose it again. So, your first choice would have 16 possibilities, and your next choice would then have 15 possibilities, then 14, 13, etc. And the total permutations would be: • 16 × 15 × 14 × 13 × ... = 20,922,789,888,000 • If you have five cards, how many ways can you line them up? • 5! = 5 x 4 x 3 x 2 x 1 = 120

  29. Problem • A student has a penny, nickel, dime, and quarter. How many different ways can she line them up? • How many different ways can these 7 Mini-Coopers be lined up? 24 5,040

  30. Permutations • If you select j objects without replacement (such as cards from a deck of cards) from a group of n objects, and you wish to count each ordering separately, then the number of ways of making the selections is given by • This formula gives the number of permutations of n objects taken j at a time, or nPj .

  31. Example • If a softball league has 10 teams, how many different end of the season rankings are possible?  (Assume no ties). • The math is n = 10, j = 10, • 10P10 = 10!/(10 − 10)! = 3,628,800 • In what order could the first 3 of 16 pool balls be in? • The math is n = 16, j = 3, • 16P3 = n!/(n − j)! = 16!/(16 − 3)! = 3,360

  32. Example • How many ways can two horses end up in the lead in a race with eight horses? • 8 possibilities for the first place horse. • 7 possibilities for the second place horse. • Number of possibilities = 8x7 =56 • Now, use the equation, with n = 8, j = 2, • nPj = n!/(n − j)! = 8P2 = 8!/6! = 8 x 7 = 56 • The probability that you will select the top two in order at random is 1/56 = 0.018 (small).

  33. Ice Cream Cones • You want variety in your ice cream cones. • Store offers 31 flavors. • You want a double dip, with two different flavors. How many options do you have? • Note that vanilla on top, chocolate on bottom is counted as a different option than chocolate on top, vanilla on bottom.

  34. Combinations • If you don’t want every possible pair of ice cream flavors counted twice, there are only 465 combinations. • If order doesn’t matter, it’s a combination. If order does matter, it’s a permutation. • If you select j objects without replacement from a group of n objects, and you do not wish to count each ordering separately, then the number of ways of making the selections is given by

  35. Example • You want a double dip, with two different flavors, but you don’t care about order, just that they have two flavors. How many options do you have? • Combination of 31 objects taken 2 at a time. • 31C2 (say “31 choose 2”) =

  36. Example • How many different committees of 4 students can be chosen from a group of 15? • 15C4 = (15!)/(4! 11!) =1,365 • In a conference of 9 schools, how many intraconference football games are played during the season if the teams all play each other exactly once? • There are 9C2 ways of choosing two teams. • 9C2 = (9!)/(2! 7!) = 36

  37. Example • You are forming a 12-member swim team from 10 girls and 15 boys. The team must consist of five girls and seven boys. How many 12-member teams are possible? • There are 10C5 of choosing five girls. There are 15C7 ways of choosing seven boys. • Using the Fundamental Counting Principle, there are 10C5• 15C7 ways choosing five girls and seven boys. • 10C5• 15C7 =(252)(6435) = 1,621,620

  38. Problem • From a group of 40 people, a jury of 12 people is to be selected. In how many different ways can the jury be selected? 5,586,853,480

  39. Problem • A university hospital hires 8 people for 4 openings in the R&D department. Three of the 8 people are women. If all 8 are qualified, in how many ways can the employer fill the four positions if • a) the selection is random? • b) exactly two selections are women? • Hint: look at example with the teams. For the women, 3 choose 2, for the men, 5 choose 2. 70 30

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