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- Arizona
- Arizona State University - Tempe
- Statistics
- Statistics 421
- Eubank
- Lecture - 02.04.09 - Equally Likely Sample Space Probability

Garrett F.

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STP421: Probability February 4, 2009 Page 1 Lecture – 02.04.09 - Equally Likely Sample Space Probability Equally Likely Sample Space Probability Proposition: 𝐸 ⊆ 𝑆 𝑃 𝐸 = 𝑛(𝐸)𝑛(𝑆) Example #1: The numbers 1,2,…,20 are written on slips of paper and placed in a hat, with 1 selected at random. 𝑛 𝑆 = 20 A) Find the probability that the drawn number is prime. 2, 3, 5, 7, 11, 13, 17, 19 𝑛 𝐸1 = 8 𝑃 𝐸1 = 820 = 25 = 0.40 B) Find the probability that the drawn number is divisible by three. 3, 6, 9, 12, 15, 18 𝑛 𝐸2 = 6 𝑃 𝐸2 = 620 = 310 = 0.30 C) Find the probability that the drawn number is either prime or divisible by three. 𝑃 𝐸1 ∪𝐸2 = 1320 = 0.65 D) Find the probability that the drawn number is both prime and divisible by three. 𝑃 𝐸1 ∩𝐸2 = 120 = 0.05 Example #2: Four people arrive randomly at a box office to buy a ticket: 𝐴,𝐵,𝐶, and 𝐷. 𝑛 𝑆 = 4! A) Find the probability that 𝐴 is the first in line. 𝑛 𝐸1 = 4!4 = 3! 𝑃 𝐸1 = 3!4! = 624 = 14 = 0.25 B) If there are only two tickets available, find the probability that 𝐴 gets a ticket. Solution 1: 𝑃 𝐸2 = 𝑃 𝐴 𝑖𝑠 𝑓𝑖𝑟𝑠𝑡 + 𝑃 𝐴 𝑖𝑠 𝑠𝑒𝑐𝑜𝑛𝑑 (since 𝐴 can’t be both) 𝑃 𝐸2 = 3!4! + 3!4! = 6+624 = 1224 = 12 = 0.5 Solution 2: There are 42 = 4!2!2! = 6 ways to fill the first two spots in line (𝑛 𝑆 ) There are 31 = 3 ways to have 𝐴 in the first two spots (𝑛 𝐸2 ) 𝑃 𝐸2 = 3 1 42 = 36 = 12 = 0.5 Example #3: The numbers in the set {1, 1, 2, 2, 2, 3, 3, 3, 4, 4} are written on slips of paper and are drawn from a hat without replacement. 𝑛 𝑆 = 10!2!3!3!2! = 25,200 A) Find the probability that all like numbers are together in the order in which they are drawn (ie: 2, 2, 2, 4, 4, 3, 3, 3, 1, 1 has all like numbers together, whereas 1, 2, 3, 4, 1, 2, 3, 4, 2, 3 does not). Since the order in which the two one’s, for example, appear does not matter, you can simply permute the sets (one’s, two’s, etc.). 𝑛 𝐸1 = 4! 𝑃 𝐸1 = 4!25,200 = 2425,200 = 11050 B) Find the probability that the two one’s are together. Let 𝑎 = 1, 1 be one element (group the one’s together) and permute the other all of the elements. 𝑛 𝐸2 = 9!1!3!3!2! = 5,040 𝑃 𝐸2 = 5,04025,200 = 15 = 0.2 STP421: Probability February 4, 2009 Page 2 A B C Example #4: There are three bars (A, B, and C) downtown as outlined in the diagram. You are at bar A and are travelling to bar B. However, you have to pick up your friend at bar C (before the cops do). Since you’ve been drinking, your route selection is completely random, though you will still take the least amount of blocks to get to bar B (6 blocks east and 4 blocks north). What is the probability that, on your way to bar B, you make it to bar C and can pick up your friend? You must travel 𝐸𝐸𝐸𝐸𝐸𝐸𝑁𝑁𝑁𝑁 blocks (in any arrangement). 𝑛 𝑆 = 10!6!4! = 210 𝐸1: Getting from bar A to bar C 𝑛 𝐸1 = 6!4!2! = 15 𝐸2: Getting from bar C to bar B 𝑛 𝐸2 = 4!2!2! = 6 𝑃 𝐸 = 6! 4!2!∗ 4! 2!2! 10! 6!4! = 15∗6210 = 90210 = 37 = 0.4286 Example #5: Suppose we want to place 10 identical objects into 4 boxes, such as in Lecture 01.26.09 Example #5. They’re filled by you and your friend from Example #4, who after closing down three bars are such that any choices you make can only be considered random. ∗⊓∗⊓∗ | ∗⊓∗ | ∗⊓∗ | ∗⊓∗⊓∗ ⨅ is considered a space between objects and | are considered dividers between boxes. Let the following describe the events: 𝐸1: No box is empty. 𝐸2: One box is empty. 𝐸3: One or more boxes is empty. Number of Empty Boxes Number of Ways to Fill the Boxes 0 9! 6! 3! = 84 1 4 1 9! 7! 2! = 4 ∗ 36 = 144 2 4 2 9! 8! 1! = 6 ∗ 9 = 54 3 4 3 9! 9! 0! = 4 ∗ 1 = 4 𝑛 𝑆 = 84 + 144 + 54 + 4 = 286 𝑃 𝐸1 = 84286 = 0.2937 𝑃 𝐸2 = 144286 = 0.5035 𝑃 𝐸3 = 1 −𝑃 𝐸1 = 1 − 84286 = 202286 = 0.7063 𝑃 𝐸3 = 144286 + 54286 + 4286 = 202286 = 0.7063 STP421: Probability February 4, 2009 Page 3 Table P1 P2 P3 P4P5P6 P7 P8 Table C1 P3 P4 P5P6 P7 P8 Table C1 C2 P5 P6 P7 P8 Example #6: There are 4 couples seated randomly around a table. Find the probability that no couple sits together (𝐸). (𝑃1,𝑃2,𝑃3,𝑃4,𝑃5,𝑃6,𝑃7,𝑃8) is the same seating as (𝑃2,𝑃3,𝑃4,𝑃5,𝑃6,𝑃7,𝑃8,𝑃1) The number of arrangements is 8!. However, each seating arrangement has seven other seating arrangements that are identical in terms of individual placement. So, 8! = 8 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑑𝑖𝑠𝑡𝑖𝑛𝑐𝑡𝑠 𝑎𝑟𝑟𝑎𝑛𝑔𝑒𝑚𝑒𝑛𝑡𝑠 → 𝑛 𝑆 = 8!8 = 7! = 5,040 Let the following describe the events: 𝐸1: Couple 1 sits together. 𝐸2: Couple 2 sits together. 𝐸3: Couple 3 sits together. 𝐸4: Couple 4 sits together. 𝐸 = 𝐸𝑖4𝑖=1 𝑃 𝐸𝑖4𝑖=1 = 𝑃 𝐸1 + 𝑃 𝐸2 + 𝑃 𝐸3 + 𝑃 𝐸4 − [𝑃 𝐸1 ∩𝐸2 + 𝑃 𝐸1 ∩𝐸3 +𝑃 𝐸1 ∩𝐸4 + 𝑃 𝐸2 ∩𝐸3 + 𝑃 𝐸2 ∩𝐸4 + 𝑃 𝐸3 ∩𝐸4 ] +[𝑃 𝐸1 ∩𝐸2 ∩𝐸3 + 𝑃 𝐸1 ∩𝐸2 ∩𝐸4 + 𝑃 𝐸1 ∩𝐸3 ∩𝐸4 +𝑃(𝐸2 ∩𝐸3 ∩𝐸4)] −𝑃(𝐸1 ∩𝐸2 ∩𝐸3 ∩𝐸4) 𝑃(𝐸1) can be seen more easily in the diagram to the right. The probability that couple one will sit together will be the number of ways to place the couple with regards to other guests multiplied by the number of ways to seat the couple itself. So, 𝑃 𝐸1 = 𝑃 𝐸2 = ⋯ = 6!2!7! = 14405040 = 27 = 0.2857 𝑃 𝐸1 ∩𝐸2 = 5!227! = 4805040 = 221 = 0.0952 𝑃 𝐸1 ∩𝐸2 ∩𝐸3 = 4!237! = 1925040 = 4105 = 0.0381 𝑃 𝐸1 ∩𝐸2 ∩𝐸3 ∩𝐸4 = 3!247! = 965040 = 0.0190 𝑃 𝐸 = 1 − 4 27 − 6 221 + 4 4105 − 965040 = 74105 = 1 − 0.7048 = 0.2952 Garrett STP421: Probability

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