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Calculating Probability of Reparation Costs for a Device with Fragile Pieces
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[QUOTE="penguin007, post: 2508709, member: 195409"] Hi everyone, I just would like to be reassured about simple probability question: [h2]Homework Statement [/h2] A device is composed by two fragile pieces during the warranty period the probability that: -none is broken is 0,25 -both are broken is 0,25 -only one is broken is 0,25 If one piece is broken then the probability that it costs -100$ is 1/2 -300$ is 1/3 -600$ is 1/6 What is the probability that the reparation costs 900$ ? [b]3. The attempt of a solution[/b] - P = ( probability that both pieces are broken that the first costs 600$, the second 300$ and the first costs 300$ and the second 600$ ) = ( 0,25 * 1/3 * 1/6 ) * 2 Do you agree with me ? Thanks! [/QUOTE]
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Calculating Probability of Reparation Costs for a Device with Fragile Pieces
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