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Rare Events, Probability and Sample Size

Rare Events, Probability and Sample Size. Rare Events An event E is rare if its probability is very small , that is, if Pr{E} ≈ 0. Rare events require large samples to ensure proper observation. Minimal Sample Size as a Function of Probability

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Rare Events, Probability and Sample Size

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  1. Rare Events, Probability and Sample Size

  2. Rare Events An event E is rare if its probability is very small, that is, if Pr{E} ≈ 0. Rare events require large samples to ensure proper observation.

  3. Minimal Sample Size as a Function of Probability Suppose that we have an Event, E, with probability PE. Then the quantity (1/ PE) represents the sample size with an expected count for E of 1. That is, expectedE≈ 1 for n ≈ (1/ PE). When PE is small, (1/ PE) is large.

  4. Rareness, Sample Size and Probability An event E is rare relative to sample size n when the expected count for E is less than 1. That is, when expectedE = n* PE < 1 n < 1/PE.

  5. Pairs of Dice and the Pair (1,1)

  6. Consider a sequence of pairs of fair dice, and the occurrence (relative to n=100) of the face-pair (1,1). Pair of Fair Dice, each with face values {1,2,3} per die (1st D3, 2nd D3) (1,1)(2,1)(3,1) (1,2)(2,2)(3,2) (1,3)(2,3)(3,3) Pr{(1,1) shows} = Pr{1 shows from 1st D3}*Pr{1 shows from 2nd D3} = (1/3)*(1/3) = 1/9  .1111111 In random samples of 100 tosses of the pair of dice, we expect approximately 100*Pr{(1,1)} = 100*(1/9)  11.11111. The smallest sample size for which we expect to observe one or more tosses showing the pair (1,1) is 1/(1/9) = 9 < 100.

  7. Consider a sequence of pairs of fair dice, and the occurrence (relative to n=100) of the face-pair (1,1). Pair of Fair Dice, one with face values {1,2,3,4} per die and one with face values {1,2,3} per die: (1st D4, 2nd D3) (1,1)(2,1)(3,1)(4,1) (1,2)(2,2)(3,2)(4,2) (1,3)(2,3)(3,3)(4,3) Pr{(1,1) shows} = Pr{1 shows from 1st D4}*Pr{1 shows from 2nd D3} = (1/4)*(1/3) = 1/12  .0833 In random samples of 100 tosses of the pair of dice, we expect approximately 100*Pr{(1,1)} = 100*(1/12)  8.33 The smallest sample size for which we expect to observe one or more tosses showing the pair (1,1) is 1/(1/12) = 12 < 100.

  8. Consider a sequence of pairs of fair dice, and the occurrence (relative to n=100) of the face-pair (1,1). Pair of Fair Dice, each with face values {1,2,3,4} per die: (1st D4, 2ndD4) (1,1)(2,1)(3,1)(4,1) (1,2)(2,2)(3,2)(4,2) (1,3)(2,3)(3,3)(4,3) (1,4)(2,4)(3,4)(4,4) Pr{(1,1) shows} = Pr{1 shows from 1st D4}*Pr{1 shows from 2nd D4} = (1/4)*(1/4) = 1/16  .0625 In random samples of 100 tosses of the pair of dice, we expect approximately 100*Pr{(1,1)} = 100*(1/16)  6.25 The smallest sample size for which we expect to observe one or more tosses showing the pair (1,1) is 1/(1/16) = 16 < 100.

  9. Consider a sequence of pairs of fair dice, and the occurrence (relative to n=100) of the face-pair (1,1). Pair of Fair Dice, each with face values {1,2,3,4,5,6,7,8} per die: (1stD8, 2ndD8) (1,1) (2,1)(3,1)(4,1)(5,1)(6,1) (7,1)(8,1) (1,2) (2,2)(3,2)(4,2)(5,2)(6,2) (7,2)(8,2) (1,3) (2,3)(3,3)(4,3)(5,3)(6,3) (7,3)(8,3) (1,4) (2,4)(3,4)(4,4)(5,4)(6,4) (7,4)(8,4) (1,5) (2,5)(3,5)(4,5)(5,5)(6,5) (7,5)(8,5) (1,6) (2,6)(3,6)(4,6)(5,6)(6,6) (7,6)(8,6) (1,7) (2,7)(3,7)(4,7)(5,7)(6,7) (7,7)(8,7) (1,8) (2,8)(3,8)(4,8)(5,8)(6,8) (7,8)(8,8) Pr{(1,1) shows} = Pr{1 shows from 1st D8}*Pr{1 shows from 2nd D8} = (1/8)*(1/8) = 1/64  .0156 In random samples of 100 tosses of the pair of dice, we expect approximately 100*Pr{(1,1)} = 100*(1/64)  1.56 The smallest sample size for which we expect to observe one or more tosses showing the pair (1,1) is 1/(1/64) = 64 < 100.

  10. Consider a sequence of pairs of fair dice, and the occurrence (relative to n=100) of the face-pair (1,1). Pair of Fair Dice, each with face values {1,2,3,4,5,6,7,8,9,10} per die: (1stD10, 2ndD10) (1,1) (2,1) (3,1) (4,1) (5,1) (6,1) (7,1) (8,1) (9,1) (10,1) (1,2) (2,2) (3,2) (4,2) (5,2) (6,2) (7,2) (8,2) (9,2) (10,2) (1,3) (2,3) (3,3) (4,3) (5,3) (6,3) (7,3) (8,3) (9,3) (10,3) (1,4) (2,4) (3,4) (4,4) (5,4) (6,4) (7,4) (8,4) (9,4) (10,4) (1,5) (2,5) (3,5) (4,5) (5,5) (6,5) (7,5) (8,5) (9,5) (10,5) (1,6) (2,6) (3,6) (4,6) (5,6) (6,6) (7,6) (8,6) (9,6) (10,6) (1,7) (2,7) (3,7) (4,7) (5,7) (6,7) (71,7) (8,7) (9,7) (10,7) (1,8) (2,8) (3,8) (4,8) (5,8) (6,8) (7,8) (8,8) (9,8) (10,8) (1,9) (2,9) (3,9) (4,9) (5,9) (6,9) (7,9) (8,9) (9,9) (10,9) (1,10) (2,10) (3,10) (4,10) (5,10) (6,10) (7,10) (8,10) (9,10) (10,10) Pr{(1,1) shows} = Pr{1 shows from 1st D10}*Pr{1 shows from 2nd D10} = (1/10)*(1/10) = 1/100  .01 In random samples of 100 tosses of the pair of dice, we expect approximately 100*Pr{(1,1)} = 100*(1/100) =1 The smallest sample size for which we expect to observe one or more tosses showing the pair (1,1) is 1/(1/100) = 100.

  11. Pair of Fair Dice, one with face values {1,2,3,4,5,6,7,8,9,10,11,12} and one with face values {1,2,3,4,5,6,7,8,9,10} Pr{(1,1) shows} = Pr{1 shows from D12}*Pr{1 shows from D10} = (1/12)*(1/10) = 1/120  0.00833 In random samples of 100 tosses of the pair of dice, we expect approximately 100*Pr{(1,1)} = 100*(1/120)  0.833. The smallest sample size for which we expect to observe one or more tosses showing the pair (1,1) is 1/(1/120) = 120 > 100.

  12. Pair of Fair Dice, each with face values {1,2,3,4,5,6,7,8,9,10,11,12}: (1st D12, 2nd D12) Pr{(1,1) shows} = Pr{1 shows from 1st D12}*Pr{1 shows from 2nd D12} = (1/12)*(1/12) = 1/144 ≈ 0.006944444 In random samples of 100 tosses of the pair of dice, we expect approximately 100*Pr{(1,1)} = 100*(1/144) ≈ . 6944444 The smallest sample size for which we expect to observe one or more tosses showing the pair (1,1) is 1/(1/144) = 144 > 100.

  13. Pair of Fair Dice, one with face values {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20} and one with face values {1,2,3,4,5,6,7,8,9,10} Pr{(1,1) shows} = Pr{1 shows from 1st D20}*Pr{1 shows from 2nd D10} = (1/20)*(1/10) = 1/200 = 0.005 In random samples of 100 tosses of the pair of dice, we expect approximately 100*Pr{(1,1)} = 100*(1/200) = .50 The smallest sample size for which we expect to observe one or more tosses showing the pair (1,1) is 1/(1/200) = 200 > 100.

  14. Pair of Fair Dice, one with face values {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24} and one with face values {1,2,3,4,5,6,7,8,9,10} Pr{(1,1) shows} = Pr{1 shows from 1st D24}*Pr{1 shows from 2nd D10} = (1/24)*(1/10) = 1/240 ≈ 0.0042 In random samples of 100 tosses of the pair of dice, we expect approximately 100*Pr{(1,1)} = 100*(1/240) ≈ .42 The smallest sample size for which we expect to observe one or more tosses showing the pair (1,1) is 1/(1/240) = 240 > 100.

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