Answer to Question #7756 in Statistics and Probability for sivagami
2012-03-27T07:26:25-04:00
.The probability that a bulb produced by a factory will fuse after certain period of time is 0.05. Find the probability that out of 5 such bulbs
a.None fuses
b.Not more than 2 fuse
c.More than 2 fuse
1
2012-03-30T07:05:45-0400
a. None fuses P(None fuses) = (1-0.05)^5 = 0.95^5 = 0.7738 & b. Not more than 2 fuse P(Not more than 2 fuse) = P(0 fuses) + P(1 fuses) + P(2 fuses) = ... P(0 fuses) = 0.95^5 P(1 fuses) = 0.05*(1-0.05)^4 P(2 fuses) = 0.05^2*(1-0.05)^3 ... = 0.95^5 + 0.05*(1-0.05)^4 + 0.05^2*(1-0.05)^3 = 0.8167. & c. More than 2 fuse P(More than 2 fuse) = 1 - P(Not more than 2 fuse) = 1-0.8167 = 0.1833.
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