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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.

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Presentation on theme: "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."— Presentation transcript:

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 P E. Then the quantity (1/ P E ) represents the sample size with an expected count for E of 1. That is, expected E ≈ 1 for n ≈ (1/ P E ). When P E is small, (1/ P E ) 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 expected E = n* P E < 1 n < 1/P E.

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 (1 st D3, 2 nd 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 1 st D3}*Pr{1 shows from 2 nd D3} = (1/3)*(1/3) = 1/9  In random samples of 100 tosses of the pair of dice, we expect approximately 100*Pr{(1,1)} = 100*(1/9)  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: (1 st D4, 2 nd 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 1 st D4}*Pr{1 shows from 2 nd 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: (1 st D4, 2 nd D4) (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 1 st D4}*Pr{1 shows from 2 nd 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)  16 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: (1 st D8, 2 nd D8) (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 1 st D8}*Pr{1 shows from 2 nd 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: (1 st D10, 2 nd D10) (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 1 st D10}*Pr{1 shows from 2 nd 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  In random samples of 100 tosses of the pair of dice, we expect approximately 100*Pr{(1,1)} = 100*(1/120)  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}: (1 st D12, 2 nd D12) Pr{(1,1) shows} = Pr{1 shows from 1 st D12}*Pr{1 shows from 2 nd D12} = (1/12)*(1/12) = 1/144 ≈ In random samples of 100 tosses of the pair of dice, we expect approximately 100*Pr{(1,1)} = 100*(1/144) ≈ 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 1 st D20}*Pr{1 shows from 2 nd D10} = (1/20)*(1/10) = 1/200 = 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 1 st D24}*Pr{1 shows from 2 nd D10} = (1/24)*(1/10) = 1/240 ≈ 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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