please solve both Example: A box contains 12 colored balls, ofwhich 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 = 5/12E₂ = black ball, with probability p₂ = 4/12E3= red ball, with probability p3= 3/12●5 4 304√(1-42-45-7, 12.12.12)-(1.2))*(*)(6X2, X3where 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 failureX₂=#breakdowns due to mechanical failureX3 = #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)

Note: Hi! Thank you for the question. The first image you have provided is simply an example of a…

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

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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