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Probability and Combinations Practice Activities

Practice solving probability problems and combinations using different scenarios and calculations. Includes bellwork exercises and mixed practice questions.

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Probability and Combinations Practice Activities

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  1. total: U7D8 pencil, red pen, highlighter, GP notebook, calculator Have out: Bellwork: Solve the following in your graph paper notebook. 1. Determine the values without using a calculator. b) a) 2. What is the probability of being dealt a heart flush in a 5–card hand? (heart flush = all hearts) 3. You have been dealt 4 cards that are all spades (♠). What is the probability that you will be dealt one more spade? Assume you are the only player for this hand.

  2. Bellwork: 1. Determine the values without using a calculator. b) a) +1 +1 +1 +1 2 2 +1 +1

  3. total: 2. What is the probability of being dealt a heart flush in a 5–card hand? (heart flush = all hearts) # of possible 5–card hands of that are all hearts. P (all hearts) = +2 Total # of possible 5–card hands +2 3. You have been dealt 4 cards that are all spades (♠). What is the probability that you will be dealt one more spade? Assume you are the only player for this hand. # of spades left in the deck 9 P (one more spade) = 48 Number of cards remaining in the deck +2

  4. IC – 82 After seven days of temperatures above 100°F, Millie’s Ice Cream Parlor had only five flavors left: chocolate fudge, French vanilla, maple-nut, lemon custard, and blueberry delight. • How many different triple-decker cones with all scoops different could they make? 5P3 = 60 b) How many different triple-decker cones could they make? 53 = 125 c) How many three-scoop dishes of ice cream with three different flavors could they serve? 5C3 = 10 d) How many three scoop dishes of ice cream were possible? nope! This is something a little different… 5C3 5P3 Not! no! 53

  5. Let’s start with an easier example: What if you are having 2 scoops with 5 flavors? Order doesn't matter; repeats are allowed. To make the problem easier, let’s temporarily assume that repeats are not allowed. Since order does not matter, divide by the arrangements. 5 ● 4 5C2 = 2! So, there are 10 combinations when there are no repeats. = 10 However, we still need to consider all the repeats of the 5 flavors of ice cream. Therefore, the total combinations are: BD, BD FV, FV LC, LC MN, MN 10 + 5 = 15 CF, CF

  6. IC – 82 After seven days of temperatures above 100°F, Millie’s Ice Cream Parlor had only five flavors left: chocolate fudge, French vanilla, maple-nut, lemon custard, and blueberry delight. d) How many three scoop dishes of ice cream were possible? There is a formula: 5+3-1C3 =7C3 n+r-1Cr

  7. IC – 83 The four questions in the preceding problem represent the four different situations that are possible when repetition and order of ice cream flavors are variables. Use the chart below and fill in a description of the problem that fits into each place along with the calculation you did. Identify the situations that are permutations or combinations. Order Important Order Not Important Decision chart “Modified” Combination n+r-1Cr Repetition Allowed d) 5+3-1C3 =7C3 53 b) Permutation nPr Combination nCr Repetition Not Allowed 5P3 a) c) 5C3

  8. Let’s make sure that your chart is complete: Does order matter? NO YES are repeats allowed? are repeats allowed? NO YES NO YES Combination Combination Decision Chart Permutation nCr n+r-1Cr nr nPr

  9. Mixed Practice 1. Suppose you draw 5 cards from a regular deck of cards. Answer part (a), then determine the probabilities for parts (b) – (j). • How many ways are there to choose 5 cards from a set of 52? b) all red P(all red)= c) all hearts d) 3 hearts and 2 spades P(all ♥)= P(3 ♥’s ,2♠’s)=

  10. f) exactly 2 diamonds (the other cards are not diamonds) e) 4 diamonds and 1 club P(4 ♦’s,1 ♣)= P(2 ♦’s,3 other)= g) exactly 3 spades h) exactly 4 face cards P(3♠’s)= P(4 face)=

  11. i) exactly 3 kings j) exactly 2 clubs, 2 diamonds P(3 kings)= P(2 ♣’s, 2 ♦’s)=

  12. 2. Seven people go to the movies. In how many ways can they arrange themselves if there are: a) 3 couples and one loner b) 2 couples and 3 others c1 c2 c3 L1 c1 c2 L1 L2 L3 4! 2! 2! 2! = 192 5! 2! 2! = 480 c) 1 couple and 5 others d) 3 boys and 4 girls that must sit next to the opposite gender (i.e. they alternate) c1 L1 L2 L3 L4 L5 6! 2! =1440 G B G B G B G Boys: 3! 4! Girls: 1 ∙3! 4! = 144 e) 2 sets of conjoined twins and 3 others L1 L2 L3 c1 c2 5! =120

  13. f) 7 people, but 2 of them hate each other and cannot sit next to each other 5 different seats for h2. h2 h2 h2 h2 h1 h2 5 • 5! 5! for all loners. 4 different seats for h2. 4 • 5! h2 h2 h2 h2 h1 5! for all loners. 4 different seats for h2. h2 h2 4 • 5! h2 h2 h1 5! for all loners. 4 different seats for h2. h2 h2 h2 h2 h1 4 • 5! 5! for all loners. h2 h2 4 different seats for h2. h2 h2 h1 4 • 5! 5! for all loners. h2 h2 4 different seats for h2. h2 h2 h1 4 • 5! 5! for all loners. 5 different seats for h2. h2 h2 h2 h2 h1 h2 5 • 5! 5! for all loners. 3600 Total= ____________

  14. Determine whether each situation involves a permutation or a combination. Then find the number of possibilities • Seating 8 students in 8 seats in the front row of the school auditorium 8P8 = 40,320 Permutation 2) Introducing the 5 starting players on the Woodsville High School basketball team at the beginning of the next basketball game Permutation 5P5 = 120 3) Checking out 3 library books from a list of 8 books for a research paper Combination 8C3 = 56 4) Choosing 2 movies to rent from 5 movies Combination 5C2 = 10

  15. 5) The first-, second-, and third- place finishers in a race with 10 contestants Permutation 10P3 = 720 6) Electing 4 candidates to a municipal planning board from a field of 7 candidates Combination 7C4 = 35 7) Choosing 2 vegetables from a menu that offers 6 vegetables choices Combination 6C2 = 15 8) An arrangement of the letters in the word rhombus Permutation = 5040 7P7 9) Selecting 2 of 8 choices of orange juice at a store Combination 8C2 = 28

  16. 10) Placing a red rose bush, a yellow rose bush, a white rose bush, and a pink bush in a row in a planter Permutation 4P4 = 24 11) Selecting 2 of 9 kittens at an animal rescue shelter Combination 9C2 = 36 12) An arrangement of the letters in the word isosceles. 9 letters; “s” 3 times; “e” 2 times = 30,240 13) How many hockey teams of 6 players can be formed from 14 players without regard to position played? Combination 14C6 = 3003 14) From a group of 10 men and 12 women, how many committees of 5 men and 6 women can be formed? Combination (10C5)(12C6) = (252)(924) = 232,848

  17. Finish the worksheet

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