Probability 3.4

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Probability 3.4

  1. 1. Counting Principles Chapter 3.4
  2. 2. Objectives <ul><li>Use the Fundamental Counting Principle to find the number of ways 2 or more events can occur </li></ul><ul><li>Find the number of ways a group of objects can be arranged in order </li></ul><ul><li>Find the number of ways to choose several objects from a group without regard to order </li></ul><ul><li>Use counting principles to find probabilities </li></ul>
  3. 3. The Fundamental Counting Principle <ul><li>If 1 event can occur in m ways and a second event can occur in n ways, the number of ways the 2 events can occur in sequence is m*n. </li></ul><ul><li>This rule can be extended for any number of events occurring in sequence. </li></ul>
  4. 4. Buying a car <ul><li>You can choose a Ford, Subaru, or Porsche </li></ul><ul><li>You can choose a small or medium car </li></ul><ul><li>You can choose red (R), purple (P), white(W) or green (G) </li></ul><ul><li>How many choices do you have? </li></ul><ul><li>3*2*3 = 18 choices </li></ul>
  5. 5. You can draw a tree diagram to show this
  6. 6. Security Systems <ul><li>The access code for a car’s security system consists of 4 digits. Each digit can be 0 through 9. </li></ul><ul><li>How many access codes are possible if each digit can be used only once and not repeated? </li></ul><ul><li>10*9*8*7 </li></ul><ul><li>= 5040 </li></ul>
  7. 7. Security Systems <ul><li>The access code for a car’s security system consists of 4 digits. Each digit can be 0 through 9. </li></ul><ul><li>How many access codes are possible if each digit can be repeated? </li></ul><ul><li>10*10*10*10 </li></ul><ul><li>= 10,000 </li></ul>
  8. 8. License Plates <ul><li>How many license plates can you make if a license plate consists of </li></ul><ul><li>Six (out of 26) letters each of which can be repeated? </li></ul><ul><li>Six (out of 26) letters each of which can not be repeated? </li></ul>
  9. 9. Permutations <ul><li>A permutation is an ordered arrangement of objects. </li></ul><ul><li>The number of different permutations of n distinct objects is n! </li></ul><ul><li>n! is read “n factorial” </li></ul><ul><li>5! = 5*4*3*2*1 </li></ul><ul><li>0! is defined as 1 </li></ul>
  10. 10. Baseball example <ul><li>How many starting lineups are possible for a 9 player baseball team? </li></ul><ul><li>9*8*7*6*5*4*3*2*1 </li></ul><ul><li>= 362,880 </li></ul><ul><li>Where is the factorial key on </li></ul><ul><li>your calculator? </li></ul>
  11. 11. Finding n P r <ul><li>Find the number of ways of forming 3-digit codes in which no digit is repeated: </li></ul><ul><li>Select the first digit </li></ul><ul><ul><li>(10 choices) </li></ul></ul><ul><li>Select the second digit </li></ul><ul><ul><li>(9 choices for each of the possible 10 first choices = 90) </li></ul></ul><ul><li>Select the third </li></ul><ul><ul><li>(8 choices for each of the 90 previous choices = 720) </li></ul></ul>
  12. 12. Permutations of n Objects taken r at a Time <ul><li>The number of permutations of n distinct objects taken r at a time is </li></ul><ul><li>n P r = n! , where r ≤ n </li></ul><ul><li>(n-r)! </li></ul><ul><li>Read this “Probability of n choose r” </li></ul>
  13. 13. Permutations of n Objects taken r at a Time <ul><li>Use the formula for the last problem </li></ul><ul><li>n P r = n! , where r ≤ n </li></ul><ul><li>(n-r)! </li></ul><ul><li>10 P 3 = 10! = 10*9*8*7*6*5*4*3*2*1 </li></ul><ul><li>(10-3)! 7*6*5*4*3*2*1 </li></ul>
  14. 14. Permutations of n Objects taken r at a Time
  15. 15. Another example . . .
  16. 16. AAAB <ul><li>How many ways can this be arranged? </li></ul><ul><li>AAAB </li></ul><ul><li>AABA </li></ul><ul><li>ABAA </li></ul><ul><li>BAAA </li></ul><ul><li>4 </li></ul>
  17. 17. AAAABBC <ul><li>How many ways can this be arranged? </li></ul><ul><li>105, but it would be tedious writing these all out, so let’s use a formula: </li></ul>
  18. 18. Distinguishable Permutations
  19. 19. Try it with AAAB: <ul><li>4! = 4*3*2*1 = 4 </li></ul><ul><li>3!*1! 3*2*1 </li></ul>
  20. 20. Try it with AAAABBC <ul><li>7! = 7*6*5 = 105 </li></ul><ul><li>4!*2!*1! 2 </li></ul>
  21. 21. Combinations <ul><li>You want to buy 3 hats from a selection of 5. How many possible choices do you have? </li></ul><ul><li>ABC ABD ABE </li></ul><ul><li>ACD ACE </li></ul><ul><li>ADE </li></ul><ul><li>BCD BCE </li></ul><ul><li>BDE CDE </li></ul>10 Choices
  22. 22. Combinations <ul><li>A combination is a selection of r objects from a group of n objects without regard to order and is denoted by n C r. </li></ul><ul><li>The number of combinations of r objects selected from a group of n objects is </li></ul><ul><li>n C r = = n! . </li></ul><ul><li>(n-r)!r! </li></ul>
  23. 23. Combinations <ul><li>Try this with the 5 hats, choose 3: </li></ul><ul><li>n C r = = n! . </li></ul><ul><li>(n-r)!r! </li></ul><ul><li>n C r = 5! . = 5*4*3*2*1 = 10 </li></ul><ul><li>(5-3)!*3! 2*1*3*2*1 </li></ul>
  24. 24. MISSISSIPPI <ul><li>A word consists of 1 M, 4 I’s, 4 S’s, and 2 P’s. If the letters are randomly arranged in order, what is the probability that the arrangement spells out Mississippi? </li></ul><ul><li>How many distinguishable permutations? </li></ul><ul><li>11! = </li></ul><ul><li>1!*4!*3!*2! </li></ul><ul><li>34,650 </li></ul><ul><li>Probability(Mississippi) = 1/34,650 = .000029 </li></ul>
  25. 25. Diamond Flush <ul><li>Find the probability of being dealt 5 diamonds from a standard deck of cards. </li></ul><ul><li>How many ways to choose 5 out of 13 diamonds? </li></ul><ul><li>13 C 5 </li></ul><ul><li>How many ways to choose a 5 card hand? </li></ul><ul><li>52 C 5 </li></ul><ul><li>P(diamond flush) = 13 C 5 = 1,287 = .0005 </li></ul><ul><li>52 C 5 = 2,598,960 </li></ul>
  26. 26. Homework <ul><li>P. 157 12-40 every 4 th problem </li></ul>
  27. 27. Now try this . . . <ul><li>With Skittles! </li></ul>

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