54046 example 3 this is geometric problem with p a p x

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Unformatted text preview: obability that all 3 are found nondefective. Therefore the probability that the lot is rejected is 1-P(accept the lot). P (accept) = 0.30 ￿4￿￿6￿ ￿1￿￿9￿ 0 3 ￿10￿ + 0.70 0 10￿ = 0.54. ￿3 3 3 So, P(reject)= 1−P(accept)=1-0.54=0.46. Example 3 This is geometric problem with p = a. P (X = n) = ( M M +N . N M M N n−1 )n−1 = . N +M N +M ( M + N )n b. P (X ≥ k ) = ∞ ￿ x=k ( N M )x−1 N +M N +M M M +N ∞ ￿ x=k ( = N )x−1 M +N = M N N N [( )k−1 + ( )k + · · · ] = ( )k−1 . M +N M +N M +N M +N Example 4 This is a binomial problem together with some probability (law of total probability and Bayes’ theorem). Let Li be the event that lot was produced on line i. Let X be the number of defective fuses among the 3 selected. a. P (L1 ∩ X = 1) P (X = 1) = P (X = 1/L1 )P (L1 ) P (X = 1/L1 )P (L1 ) + P (X = 1/L2 )P (L2 ) + P (X = 1/L3 )P (L3 ) + P (X = 1/L4 )P (L4 ) + P (X = 1/L5 )P (L5 ) = P (L1 /X = 1) = ￿ 3￿ 1 0.051 0.952 1 + 5 ￿ 3￿ 1 ￿ 3￿ 0.051 10.952 1 1 5 ￿3￿ ￿ 3￿ 1 0.021 0.982 1 + 5 1 0.021 0.982 5 + b. The probability that the lot came from one of the other 4 lines is: P (L2 ∪ L3 ∪ L4 ∪ L5 /X = 1) = 1 − P (L1 /X = 1) = 1 − 0.37 = 0.63. 15 1 0.021 0.982 1 + 5 ￿ 3￿ 1 0.021 0.982 1 5 P (L1 /X = 1) = 0.37. ⇒ Example 5 Let X be the number of tests needed for each group of 10 people. Then, if nobody has the disease 1 test is enough. But if the test is positive then there will be 11 test (1 + 10). The probability distribution of X is: X P (X ) ￿10￿ 1 0.100 0.9010 = 0.9010 ￿ ￿0 11 1 − 10 0.100 0.9010 = 1 − 0.9010 0 Therefore the expected number of tests is: 10 + 11(1 − 0.9010 = 7.51. E (X ) = 1(0.90) Example 6 This is a hypergeometric problem. Let’s define the events: D: Selected card is ♦ (part a) H 4: All 4 selected cards are ♥ (part b) Lc : Lost card is ♣ Ls : Lost card is ♠ Lh : Lost card is ♥ Ld : Lost card is ♦ a. We need to find P (Lc /D). This equal to: P (Lc /D) = P (Lc ∩ D) P (D ) P (Lc ∩ D) P (Lc ∩ D) + P (Ls ∩ D) + P (Lh ∩ D) + P (Ld ∩ D) P (D/Lc )P (Lc ) P (D/Lc )P (Lc ) + P (D/Ls )P (Ls ) + P (D/Lh )P (Lh ) + P (D/Ld )P (Ld ) 13 1 51 4 + 13 1 51 4 13 1 + 13 1 51 4 51 4 + 12 1 51 4 = = = = 13 = 0.2549. 51 b. We need to find P (Lc /H 4). This equal to: P (Lc /H 4) = P (Lc ∩ H 4) P (H 4) P (Lc ∩ H 4) P (Lc ∩ H 4) + P (Ls ∩ H 4) + P (Lh ∩ H 4) + P (Ld ∩ H 4) P (H 4/Lc )P (Lc ) P (H 4/Lc )P (Lc ) + P (H 4/Ls )P (Ls ) + P (H 4/Lh )P (Lh ) + P (H 4/Ld )P (Ld ) ￿13￿￿38￿ 4 ￿51￿0 1 4 4 ￿13￿￿38￿ ￿13￿￿38￿ ￿12￿￿39￿ ￿13￿￿38￿ 4 ￿51￿0 1 + 4￿51￿0 1 + 4￿51￿0 1 + 4￿51￿0 1 4 4 4 4 4 4 ￿13￿ 4 4 + ￿13￿ ￿13￿ 4 ￿12￿ 4 + 4 4 + ￿13￿ = 0.2708. 4 Example 7 Here, X takes on two values 0 or 1. Let P (X = 1) = p. The probability distribution of X is: X P (X ) 0 1-p 1 p a. E (X ) = 0(1 − p) + 1(p) ⇒ E (X ) = p. V ar(X ) = 02 (1 − p) + 12 (p) − P 2 ⇒ V ar(X )...
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This note was uploaded on 10/04/2012 for the course STATISTICS 100a taught by Professor Cristou during the Spring '10 term at UCLA.

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