# SOLUTION: A production line involves the useof three machines consecutively.The chance that the first machine will breakdown in any one week is 0.1,for the second machine to breakdown is 0.0

Algebra ->  Algebra  -> Probability-and-statistics -> SOLUTION: A production line involves the useof three machines consecutively.The chance that the first machine will breakdown in any one week is 0.1,for the second machine to breakdown is 0.0      Log On

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 Click here to see ALL problems on Probability-and-statistics Question 260675: A production line involves the useof three machines consecutively.The chance that the first machine will breakdown in any one week is 0.1,for the second machine to breakdown is 0.05 and for the third to breakdown is 0.025.What is the probability that(a)there will be no breakdown? b)at least vtwo of the machines break down in a certain week?Answer by Theo(3458)   (Show Source): You can put this solution on YOUR website!p(first machine break down) = .1 p(second machine break down) = .05 p(third machine break down) = .025 p(first machine will not break down = 1 - .1 = .9 p(second machine will not break down = 1 - .05 = .95 p(third machine will not break down = 1 - .025 = .975 the probability that no machines will break down = .9 * .95 * .975 = .833625 the probability that all 3 will break down = .1 * .05 * .025 = .000125 the probability that the first machine will break down only = .1 * .95 * .975 = .092625 the probability that the second machine will break down only = .9 * .05 * .975 = .043875 the probabillity that the third machine will break down only = .9 * .95 * .025 = .021375 the probability that the first and second machine break down = .1 * .05 * .975 = .004875 the probability that the first and third machine break down = .1 * .95 * .025 = .002375 the probability that the second and third machine break down = .9 * .05 * .025 = .001125 total probability is 1 as it should be. you had to do each scenario separately because the probability for each occurrence was different. if they were the same, you could have used combination formulas as follows: assume probability for each of the machines to break down was the same. formula then would have been as follows: p(0) = .9^3 * 1 = .729 p(1) = .1^1 * .9^2 * 3 = .243 p(2) = .1^2 * .9^1 * 3 = .027 p(3) = .1^3 * 1 = .001 total probability = 1 as it should be. multiplication factors were based on combination formulas as follows: number of possible ways to get 0 occurrences = c(3,0) = 3! / 0!*3! = 1 number of possible ways to get 1 occurrence = c(3,1) = 3! / 1!*2! = 3 number of possible ways to get 2 occurrences = c(3,2) = 3! / 2!*1! = 3 number of possible ways to get 3 occurrences = c(3,3) = 3! / 3!*0! = 1 your problem was more complex because each of the machines had a different probability of breaking down which is why each of the machines had to be treated differently. you had 8 scenarios in each problem. with your more complex problem you had to identify each of the scenarios uniquely because the probabilities for each were unique. with my more simple problem you were able to combine 3 scenarios together for 1 occurrence and 2 occurrences because the probabilities were the same for all 3 machines. this is how the different scenarios work out for both your complex problem and my simpler problem. ``` breakdowns machine 1 machine 2 machine 3 0 n n n 1 y n n 1 n y n 1 n n y 2 y y n 2 y n y 2 n y y 3 y y y ```