misuse What is the probability of 2 electrical failures, 4 mechanical failures, and 3 misuses? 79 x₁,x₂,3; 9,0.2, 0.5,0.3)= (0.2)² (0.5)¹ (0.3)³ 2,4,3
misuse What is the probability of 2 electrical failures, 4 mechanical failures, and 3 misuses? 79 x₁,x₂,3; 9,0.2, 0.5,0.3)= (0.2)² (0.5)¹ (0.3)³ 2,4,3
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
Related questions
Question
![Machine Breakdown Example
• S={E= electrical, M = mechanical, U = misuse}
●
• P(E)= 0.2, P(M)= 0.5, P(U) = 0.3
●
Pay
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,.,;:9,02,05,03)-(243)(02) (05) (0.3)”
17](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fcd1827cf-4955-4776-9665-11d8be2adcd3%2F636ab882-6b6d-400a-9825-006f9afc5044%2F3u2k3i_processed.jpeg&w=3840&q=75)
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
●
Pay
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,.,;:9,02,05,03)-(243)(02) (05) (0.3)”
17
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