Machine Breakdown Example • S={E= electrical, M = mechanical, U = misuse} P(E)= 0.2, P(M) = 0.5, P(U) = 0.3 ● X₁ = #breakdowns due to electrical failure X₂ = #breakdowns due to mechanical failure X3 = =#breakdowns due to operator misuse ➤What is the probability of 2 electrical failures, 4 mechanical failures, and 3 misuses? ƒ(x₁,x₂,.x3;9,0.2,0.5,0.3) = 9 (0.2) (0.5)*(0.3)³ 2,4,3 17

A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
Section: Chapter Questions
Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
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Example: A box contains 12 colored balls, of
which 5 are white, 4 are black, and 3 are red.
Sample 7 balls with replacement.
• n = 7 trials, k= 3 possible outcomes
●
●
E₁= white ball, with probability p₁ = 5/12
E₂ = black ball, with probability p₂ = 4/12
E3= red ball, with probability p3= 3/12
●
5 4 3
0
4
√(1-42-45-7, 12.12.12)-(1.2))*(*)
(6
X2, X3
where x₁ + x₂ + x3 = 7
Transcribed Image Text:Example: A box contains 12 colored balls, of which 5 are white, 4 are black, and 3 are red. Sample 7 balls with replacement. • n = 7 trials, k= 3 possible outcomes ● ● E₁= white ball, with probability p₁ = 5/12 E₂ = black ball, with probability p₂ = 4/12 E3= red ball, with probability p3= 3/12 ● 5 4 3 0 4 √(1-42-45-7, 12.12.12)-(1.2))*(*) (6 X2, X3 where x₁ + x₂ + x3 = 7
Machine Breakdown Example
• S={E= electrical, M = mechanical, U = misuse}
●
P(E)= 0.2, P(M) = 0.5, P(U) = 0.3
●
X₁ = #breakdowns due to electrical failure
X₂=#breakdowns due to mechanical failure
X3 = #breakdowns due to operator misuse
➤What is the probability of 2 electrical failures,
4 mechanical failures, and 3 misuses?
ƒ (x₁,x₂, x3; 9,0.2,0.5,0.3) =
9
(2.4.3) (02)*(0.5) (0.3)
Transcribed Image Text:Machine Breakdown Example • S={E= electrical, M = mechanical, U = misuse} ● P(E)= 0.2, P(M) = 0.5, P(U) = 0.3 ● X₁ = #breakdowns due to electrical failure X₂=#breakdowns due to mechanical failure X3 = #breakdowns due to operator misuse ➤What is the probability of 2 electrical failures, 4 mechanical failures, and 3 misuses? ƒ (x₁,x₂, x3; 9,0.2,0.5,0.3) = 9 (2.4.3) (02)*(0.5) (0.3)
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