Rubik’s Cube

1. If you have a small bag of skittles and you separate then, you have 5 blue, 6 red, 3 purple, and 5 yellow. What is the probability that you would pick a blue skittle?

2. Rubik’s Cube • First developed in Hungary in the 1970’s bu Erno Rubik, a rubik’s cube contains 26 small cubes. The square faces of the cubes are colored in six different colors. The cubes can be twisted horizontally or vertically. When first purchased, the cube is arranged so that each face shows a single color. To do the puzzle, you first turn columns and rows in a random way until all of the six faces are multicolored.

3. To solve the puzzle, you must return the cube to its original state – that is a single color on each of the six faces. With 115,880,067,072,000 arrangements, this is no easy task. If it takes one-half second for each of these arrangements, it would require over 1,800,000 years to move the cube into all possible arrangements.

4. Thinking Mathematically I can compute probabilities with permutations. I can compute probabilities with combinations.

5. Example Probability and Permutations Five groups in a tour, Lady GaGa, Aerosmith, Oak Ridge Boys, the Rolling Stones, and the Beatles, agree to determine the order of performance based on a random selection. Each band’s name is written on one of five cards. The cards are placed in a hat and then five cards are drawn out, one at a time. The order in which the cards are drawn determines the order in which the bands perform. What is the probability of the Rolling Stones performing fourth and the Beatles last?

6. Solution We begin by applying the definition of probability to this situation. P(Rolling Stones fourth, Beatles last) = (permutations with Rolling Stones fourth, Beatles last) (total number of possible permutations) We can use the Fundamental Counting Principle to find the total number of possible permutations. 5  4  3  2  1 = 120

7. Solution cont. We can also use the Fundamental Counting Principle to find the number of permutations with the Rolling Stones performing fourth and the Beatles performing last. You can choose any one of the three groups as the opening act. This leaves two choices for the second group to perform, and only one choice for the third group to perform. Then we have one choice for fourth and last. 3  2  1  1  1 = 6 There are six lineups with Rolling Stones fourth and Beatles last.

8. Solution cont. Now we can return to our probability fraction. P(Rolling Stones fourth, Beatles last) = (permutations with Rolling Stones fourth, Beatles last) (total number of possible permutations) = 6/120 = 1/20 The probability of the Rolling Stones performing fourth and the Beatles last is 1/20.

9. Blitzer Bonus • Florida’s lottery game, LOTTO, is set up so each player chooses six different numbers from 1 to 53. If the six numbers chosen match the six numbers drawn randomly, the player wins (or shares) the top cash prize. That prize could range from $7 million to $106.5 million). With one LOTTO ticket, what is the probability of winning the prize?

10. P(winning) = • P(winning) = • P(winning) =

11. Try One • People lose interest when they do not win at a game of chance, including Florida’s LOTTO. With drawings twice weekly instead of one, the game described in the last example was added to bring back lost players and increase ticket sales. The original LOTTO was set up so that each player chose six different numbers from 1 to 49, rather than from 1 to 53, with a lottery drawing only once a week. With one LOTTO ticket, what was the probability of winning the top cash prize in Florida’s original LOTTO? Express the answer as a fraction and as a decimal correct to ten places.

12. Example Probability and Combinations A club consists of five men and seven women. Three members are selected at random to attend a conference. Find the probability that the selected group consists of: • three men. • one man and two women.

13. Solution We begin with the probability of selecting three men. P( 3 men)=number of ways of selecting 3 men total number of possible combinations 12C3 = 12!/((12-3)!3!) = 220 5C3 = 5!/((5-3)!(3!)) = 10 P(3 men) = 10/220 = 1/22

14. Solution part b We set up the fraction for the probability that the selected group consists of one man and two women. P(1 man and 2 women) = number of ways of selecting 1 man and 2 women total number of possible combinations We know the denominator is 12C3 = 220. Next we move to the numerator of the probability fraction.

15. Solution part b cont. The number of ways of selecting r = 1 man from n = 5 men is 5C1 = 5!/(((5-1)!1!) = 5 The number of ways of selecting r = 2 women from n=7 women is 7C2 = 7!/((7-2)!2!) = 21

16. Solution part b cont. By the Fundamental Counting Principle, the number of ways of selecting 1 man and 2 women is 5C1 7C2 = 5  21 = 105 Now we can fill in the numbers in our probability fraction. P(1 man and 2 women) = number of ways of selecting 1 man and 2 women total number of possible combinations = 105/220 = 21/44

17. You try • A club consists of six men and four women. Three members are selected at random to attend a conference. Find the probability that the selected group consists of a. Three men b. Two men and one woman

18. Thinking Mathematically Probability with the Fundamental Counting Principle, Permutations, and Combinations