A piece of equipment will function only all components A, B and C are working. The probability of A failing during one year is 1/6, that of B failing in 1/20 & that of C failing in 1/10. What is the probability that the equipment will fail before the end of the year?

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Arun · Answer

Dear Sam A will be working P(A) = 1 – 1/6 = 5/6 B ….....................P(B) = 1 – 1/20 = 19/20 C..........................P(C) = 1 – 1/10 = 9/10 Hence equipment will fail, its probability will be = 1 – [5/6 * 19/20 * 9 / 10] = 1 – [57 /80] = 23 / 80 Regards Arun

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A piece of equipment will function only all components A, B and C are working. The probability of A failing during one year is 1/6, that of B failing in 1/20 & that of C failing in 1/10. What is the probability that the equipment will fail before the end of the year?

A piece of equipment will function only all components A, B and C are working. The probability of A failing during one year is 1/6, that of B failing in 1/20 & that of C failing in 1/10. What is the probability that the equipment will fail before the end of the year?

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