Probability question?
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11-09-2012, 01:39 PM
Post: #1
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Probability question?
So this probability question was posted in a facebook group and I just can't figure it out. Anyone care to explain how it's solved? I've done some basic probability, so I'm not clueless on the subject, I just have don't this type of problem before.
The probability that an electronic device produced by a company does not function properly is equal to 0.1. If 10 devices are bought, then the probability, to the nearest thousandth, that 7 devices function properly is ??? Ads |
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11-09-2012, 01:48 PM
Post: #2
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if one device has a working to not working ratio of 0.9 : 0.1 Then
10 devices and if 7 devices have to function and three devices don't function then (0.1)^3 * (0.9)^7 = 4.782 * 10^-4 Ads |
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11-09-2012, 01:48 PM
Post: #3
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It's from the Binomial Distribution
N=10 p = 0.1 = P(doesn't work) P(X=0) = P(none of them work out of the 10) = 0.1^10 P(X=10) = P(all of them work out of the 10) = 0.9^10 P(X=3) = P(3 of them do not work) = P(7 of them work) = (ways to choose 3) * (probability the first three don't work) * (probability the last 7 work) = (0.1)^3 * (0.9)^7 * (10 choose 3) = 0.05739 10 choose 3 is 120 meaning the pattern of W (working) and N (non-working) devices could be in 120 different patterns. |
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11-09-2012, 01:48 PM
Post: #4
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probability that device doesn't work properly = 0.1
hence, probability that a device works properly = 1 - 0.1 = 0.9 now, it is given that 7 out of 10 devices function properly but you don't know which 7 devices works properly. So, you have to select randomly 7 devices out of 10 in ^10C_7 ways = 10!/(7! 3!) = 120 ways required probability = 120 x (0.9)^7 x (0.1)^3 = 0.0574 = 0.057 |
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