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Combinations, Permutations & Counting Principle

Combinations, Permutations & Counting Principle . Continued. It’s time to combine & expand. From a group of 7 men and 6 women, five people are to be selected to form a committee so that at least 3 men on the committee. In how many ways can it be done ? What are the possible scenarios

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Combinations, Permutations & Counting Principle

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  1. Combinations, Permutations & Counting Principle Continued

  2. It’s time to combine & expand • From a group of 7 men and 6 women, five people are to be selected to form a committee so that at least 3 men on the committee. In how many ways can it be done? • What are the possible scenarios • 3 men and 2 women, 4 men and 1 woman, or 5 men • (7C3 x 6C2) + (7C4 x 6C1) + (7C5) - but why do we add here? • (525 + 210 + 21) • 756

  3. MORE EXAMPLES • In how many different ways can the letters of the word 'LEADING' be arranged in such a way that the vowels always come together? • The word 'LEADING' has 7 different letters. • But, when we pair all the vowels together (EAI) they count as 1 letter • So there are 5 letters → 5! = 120 combinations • But the vowels can also be rearranged, so → 3! = 6 • When we combine these two we get 120 x 6 = 720 possible solutions

  4. Continued • In how many different ways can the letters of the word 'DETAIL' be arranged in such a way that the vowels occupy only the odd positions? • There are 6 letters in the given word, out of which there are 3 vowels and 3 consonants. • Let us mark these positions as under: • (1) (2) (3) (4) (5) (6) • Now, 3 vowels can be placed at any of the three places marked 1, 3, 5. • Number of ways of arranging the vowels = 3P3 = 3! = 6. • Also, the 3 consonants can be arranged at the remaining 3 positions. • Number of ways of these arrangements = 3P3 = 3! = 6. • Total number of ways = (6 x 6) = 36.

  5. Probabilities from two way tables Stu Staff Total American 107 105 212 European 33 12 45 Asian 55 47 102 Total 195 164 359 12) What is the probability that the driver is a student?

  6. Probabilities from two way tables Stu Staff Total American 107 105 212 European 33 12 45 Asian 55 47 102 Total 195 164 359 13) What is the probability that the driver drives a European car?

  7. Probabilities from two way tables Stu Staff Total American 107 105 212 European 33 12 45 Asian 55 47 102 Total 195 164 359 14) What is the probability that the driver is staff and drives an Asian car?

  8. Probabilities from two way tables Stu Staff Total American 107 105 212 European 33 12 45 Asian 55 47 102 Total 195 164 359 15) What is the probability that the driver drives an American or Asian car? Disjoint?

  9. Probabilities from two way tables Stu Staff Total American 107 105 212 European 33 12 45 Asian 55 47 102 Total 195 164 359 16) What is the probability that the driver is staff or drives an Asian car? Disjoint?

  10. Probabilities from two way tables Stu Staff Total American 107 105 212 European 33 12 45 Asian 55 47 102 Total 195 164 359 17) If the driver is a student, what is the probability that they drive an American car? Condition

  11. Probabilities from two way tables Stu Staff Total American 107 105 212 European 33 12 45 Asian 55 47 102 Total 195 164 359 18) What is the probability that the driver is a student if the driver drives a European car? Condition

  12. Example 19: Management has determined that customers return 12% of the items assembled by inexperienced employees, whereas only 3% of the items assembled by experienced employees are returned. Due to turnover and absenteeism at an assembly plant, inexperienced employees assemble 20% of the items. Construct a tree diagram or a chart for this data. What is the probability that an item is returned? If an item is returned, what is the probability that an inexperienced employee assembled it? P(returned) = 4.8/100 = 0.048 P(inexperienced|returned) = 2.4/4.8 = 0.5

  13. Only 5% of male high school basketball, baseball, and football players go on to play at the college level. Of these, only 1.7% enters major league professional sports. Of the athletes that do not play college sports, only 0.1% enters major league professional sports. What is the probability that a high school athlete will play professional sports? What is the probability that a high school athlete does not play college sports if he plays professional sports? 1.7% of 50 Make up a population size! 5% of 1000 P(play pro) = P(play college & Play pro) or P(not play college & play pro) = .05(.017) + (.95)(.001) = .0018 P(not play college | plays pro) = P(not play college & play pro) / P(play pro) = .95/1.8 = .5278

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