1 / 14

Worksheet 6 Answers

Worksheet 6 Answers. Problem 1. P(ace) = 4/52 There are four 4’s out of 52 cards P(diamond) = 13/52 There are 13 diamonds P(ace of diamonds) = 1/52 There is only one ace of diamonds P(4 or 6) = 4/52 + 4/52 = 8/52 These are disjoint events

candid
Download Presentation

Worksheet 6 Answers

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Worksheet 6 Answers

  2. Problem 1 • P(ace) = 4/52 There are four 4’s out of 52 cards • P(diamond) = 13/52 There are 13 diamonds • P(ace of diamonds) = 1/52 There is only one ace of diamonds • P(4 or 6) = 4/52 + 4/52 = 8/52 These are disjoint events • P(4 or club) = 4/52 + 13/52 – 1/52 = 16/52 Subtract overlap • P(red queen) = 2/52 • There are two red queens OR: • P(red AND queen) = P(red) x P(queen GIVEN red) = 26/52 x 2/26 • P(black and 10) = 2/52 Same argument • P(4 and 5) = 0 P(impossibility) = 0

  3. Problem 2 • P(red) = 5/10 • P(green) = 3/10 • P(red or white) = 5/10 + 2/10 = 7/10 Disjoint • P(red AND white) = 0 One marble can’t be two colors • P(not green) = 1 – P(green) = 1 – 3/10 = 7/10

  4. Problem 3 • P(both diamonds) = 13/52 x 13/52 = 1/16 Independent so multiply • P(same card) = 52/52 x 1/52 = 1/52 • P(both 3’s) = 4/52 x 4/52 = 1/169

  5. Problem 4 • P(both diamonds) = 13/52 x 12/51 = .059 • P(d AND d) = P(d) x P(d GIVEN d) • P(both 3’s) = 4/52 x 3/51 = .005 • P(a pair) = 52/52 x 3/51 = 3/51 = .059

  6. Problem 5 • P(A OR dress) = 37/118 + 54/118 – 24/118 = 67/118 • P(B or C) = 54/118 + 27/118 = 81/118 • P(blouse or A) = 64/118 + 37/118 – 13/118 = 88/118

  7. Problem 6 • P(alcoholic and elevated) = 87/300 • OR: 100/300 x 87/100 • P(nonalcoholic) = 200/300 • P(nonalcoholic and normal) = 157/300 • OR: 200/300 x 157/200 • You can also get these answers with a tree diagram. Assign each branch its probability and multiply.

  8. Problem 7 • P(m or p) = P(m) + P(p) – P(m and p) • .45 = .32 + .17 – x • x = .04

  9. Problem 8 • Assume that a red ball was actually drawn. Then it came from box 1 OR from box 2. These are disjoint events, so add: • P(box 1 and red GIVEN box 1) = ½ x 2/3 = 1/3 • P(box 2 and red GIVEN box 2 ) = ½ x ¼ = 1/8 • So P(red) = 1/3 + 1/8 = 11/24

  10. Problem 9 • P(b and w) = P(b) x P(w given b) • 15/56 = 3/8 x P • P = 5/7 • Note: The worksheet had a typo; it said that P(b and w) is 15/16; it should be 15/56

  11. Problem 10 • There are 50 females, so P(y given f) = 8/50 • There are 60 no’s,so P(m given no) = 18/60

  12. Problem 11 • The questions asks about “at least” probability, so find the complement: • What is the probability that NO ace is drawn? • P(no ace) = 48/52 x 47/51 x 46/50 x 45/49 • This evaluates to about 0.72 • P(at least one ace) = 1 – 0.72 = 0.28 • Note: This can be solved affirmatively with a complicated set of “and” and “or” calculations. Note, for example, that there is more than one way to get three aces, etc.

  13. Problem 12 • P(at least one shared birthday) = 1 – P(no shared birthdays) • So how can we find the probability that there are no shared birthdays? • Assume the 37 people enter the room one at a time • The first person HAS a birthday, with probability 365/365 • The next person has a probability of 364/365 of having a different birthday • The next has a probability of 363/365 of not matching the first two people, an so on. • This results in a group of 37 factors • Notice these are independent AND probabilities, so we multiply

  14. Evaluate the 37 factors:

More Related