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Aim: How do we use permutations and combinations to solve probability problems? Do Now: Six students are arranged at random on a bench. What is the probability.

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Presentation on theme: "Aim: How do we use permutations and combinations to solve probability problems? Do Now: Six students are arranged at random on a bench. What is the probability."— Presentation transcript:

1 Aim: How do we use permutations and combinations to solve probability problems?
Do Now: Six students are arranged at random on a bench. What is the probability that Ed, one of the students, is in the first seat if the bench sits six.

2 Permutations & Probability problems
Six students are arranged at random on a bench. What is the probability that Ed, one of the students, is in the first seat if the bench sits six. W/O permutations P(Ed in first seat) = 1/6 With permutations 5P5 Number of ways other 5 seats are occupied 1P1 Number of ways for Ed to side in 1st seat 5P5 6P6 = 1/6 1P1 • 6P6 Number of ways 6 seats could be occupied

3 Model Problem Two cards are drawn at random from a standard deck of 52 cards, without replacement. What is the probability that both cards drawn are fives? Dependent Events P(first 5) = 4/52 P(second 5) = 3/51 P(5, 5) = 4/52 • 3/51 = 1/221 Number of ways to draw 2 5’s from possible 4 Number of ways to draw any 2 cards from deck of 52 4C2 52C2 4C2 52C2 = 1/221

4 Probability Involving Combinations
Two cards are drawn at random from a standard deck of 52 cards, without replacement. What is the probability that both cards drawn are fives? Dependent Events 1 Counting Principle P(1st 5) = 4/52 P(2d 5) = 3/51 P(5, 5) = 4/52 • 3/51 2 Permutations 4P2 52P2 = 1/221 3 Combinations 4C2 52C2 = 1/221 Number of ways to draw 2 5’s from possible 4 Number of ways to draw any 2 cards from deck of 52 4C2 52C2

5 Permutation or Combination?
Model Problem In a school organization, there are 4 sophomores and 5 juniors. A committee of 4 people is to be selected from this group. What is the probability that 2 sophomores and 2 juniors will be on the committee? Permutation or Combination? n(combination of 2s & 2j) P(2 s, 2 j) = n(4-member combinations made from the 9 members)

6 Nine total students from which to choose
Model Problem In a school organization, there are 4 sophomores and 5 juniors. A committee of 4 people is to be selected from this group. What is the probability that 2 sophomores and 2 juniors will be on the committee? Total Outcomes n(S): Nine total students from which to choose : 9C4 = 126 Successful Outcomes n(E): How do we determine n(2 soph., 2 jun.)? Two of four sophomores 4C2 = 6 Two of five juniors 5C2 = 10 n(2 soph., 2 jun.) = 4C2 • 5C2 = 6•10 = 60

7 Model Problem In a school organization, there are 4 sophomores and 5 juniors. A committee of 4 people is to be selected from this group. What is the probability that 2 sophomores and 2 juniors will be on the committee? 4C2 • 5C2 P(2 sophs., 2 juniors) = 9C4 = 6 • 10 126 = 60 126 = 10 21

8 Successful Outcomes n(E)
Model Problem An urn contains 4 white marbles and 5 blue marbles, all of equal size. Three marbles are drawn at random with no replacement. What is the probability that at least 2 marbles drawn are blue? Total Outcomes n(S) : 9C3 Successful Outcomes n(E) At least 2 are blue: (2-b, 1-w) or (3-b) P(A  B) = P(A) + P(B) - P(A  B) 5C2 4C1 5C3 Successful Outcomes n(E) 5C2 • 4C1 = + 5C3 9C3 Total Outcomes n(S) P(at least 2 b) = 84 = 25/42

9 Model Problem A candy dish contains 10 candies. Three candies are covered with red foil and 7 with green foil. A. If 2 candies are chosen at random from the dish, what is the probability that both will be covered with the same colored foil? P(both red or both green) = P(2R) + P(2G) P(A  B) = P(A) + P(B) - P(A  B) P(2R) = 3C2 P(2G) = 7C2 10C2 10C2 P(2R or 2G) = 10C2 3C2 + 7C2 = 45 = 8/15

10 Model Problem A candy dish contains 10 candies. Three candies are covered with red foil and 7 with green foil. B. If 2 candies are chosen at random from the dish, what is the probability that each will be covered with a different foil?

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