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String of lightbulbs

Expert replies
by jfranco23 » Tue Mar 17, 2009 5:06 pm
A string of lightbulbs is wired in such a way that if any individual lightbulb fails, the entire string fails. If for each individual lightbulb the probability of failing during time period T is 0.06, what is the probability that the string of lightbulbs will fail during time period T?

A. 0.06
B. (0.06)^10
C. 1 - (0.06)^10
D. (0.94)^10
E. 1 - (0.94)^10
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Source: — Problem Solving |

by cubicle_bound_misfit » Wed Mar 18, 2009 7:52 am
whats the no of lightbulbs in the string?
Cubicle Bound Misfit
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by rseeker2 » Wed Mar 18, 2009 8:49 am
probability that the string of lightbulbs will fail during time period T --> 1-P(no bulb failing)

= 1 - (0.94)^n

If n were 10, it would be 1 - (0.94)^10
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by Stuart@KaplanGMAT » Wed Mar 18, 2009 11:00 am
rseeker2 wrote:probability that the string of lightbulbs will fail during time period T --> 1-P(no bulb failing)

= 1 - (0.94)^n

If n were 10, it would be 1 - (0.94)^10
Perfect!

Based on the answer choices, there must be 10 lightbulbs - the OP must have missed part of the question (maybe it really started with "A string of 10 lightbulbs...").
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by jfranco23 » Wed Mar 18, 2009 11:58 am
So sorry for my mistake, there are 10 lightbulbs.
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by hijazim » Thu Mar 19, 2009 7:09 am
Shouldn't it be (0.06)x10 because it is OR not AND probability. As if any of the lightbulbs failed the whole string will fail, so if the 1st or 2nd or ... 10th fail the wire will fail..

so it has to be : 0.06 + 0.06 + 0.06 + 0.06 + 0.06 + 0.06 + 0.06 + 0.06 + 0.06 + 0.06 = 0.6

please advise.
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by sureshbala » Thu Mar 19, 2009 7:42 am
hijazim wrote:Shouldn't it be (0.06)x10 because it is OR not AND probability. As if any of the lightbulbs failed the whole string will fail, so if the 1st or 2nd or ... 10th fail the wire will fail..

so it has to be : 0.06 + 0.06 + 0.06 + 0.06 + 0.06 + 0.06 + 0.06 + 0.06 + 0.06 + 0.06 = 0.6

please advise.
Folk in addition to this the light will fail if 1st AND 2nd fail and the other don't fail. 1st, 2nd AND 3rd fail and the rest don't....so on...

So instead of considering all this we calculate the probability that none of the light fails for which the probability is (0.94)^10.

Hence the probability that at least one of them fails = 1- (0.94)^10
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