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A string of 5 different colored light bulbs is wired in

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A string of 5 different colored light bulbs is wired in such a way that if any two-consecutive light bulb fails, then entire string fails.
If for each individual light bulb, the probability of not failing during time period T is 0.85, what is the probability that the string of light bulbs will fail during the time period T?

Options
a) 0.0225
b) 0.09
c) 0.225
d) 0.25
e) 0.5

OA B

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by swerve » Thu Aug 29, 2019 2:14 pm
We have to find the probability of fail, on the other hand, we have given that if two consecutive bulbs fail then only the complete string will fail.
So, given probability of not fail \(= 0.85\)

So
probability of fail \(= 0.15\)

In string of 5 in how many ways 2 consecutive will fail \(= 4\) ways (12,23,34,45)

probability of fail \(= 0.15\cdot 0.15 \cdot 4\)
\(= 0.09 \Rightarrow\) __B__

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by Ian Stewart » Fri Aug 30, 2019 7:40 am
BTGmoderatorDC wrote:A string of 5 different colored light bulbs is wired in such a way that if any two-consecutive light bulb fails, then entire string fails.
If for each individual light bulb, the probability of not failing during time period T is 0.85, what is the probability that the string of light bulbs will fail during the time period T?

Options
a) 0.0225
b) 0.09
c) 0.225
d) 0.25
e) 0.5

OA B

Source: e-GMAT
They do not have the correct answer among the answer choices (the correct answer is roughly 0.0794), and this question is way too complicated for the GMAT. I googled the question, and looked at the official explanation posted here. There they essentially just add the probabilities of four cases: bulbs 1+2 fail, or bulbs 2+3 fail, or bulbs 3+4 fail, or bulbs 4+5 fail. So they arrive at the answer (4)(3/20)^2 = 0.09. That is wrong, because the cases overlap: when bulbs 1 and 2 fail, it is sometimes true that bulb 3 also fails. If you just add the probability that 1+2 fail and the probability that 2+3 fail, you're counting the situation when bulbs 1+ 2+3 all fail twice instead of once. You simply can't add probabilities for different cases when cases overlap, unless you account for that overlap.

It's easy to see why their method is wrong just by imagining that each bulb fails 100% of the time. Using the method in the "official explanation", you'd find that the string of bulbs fails 400% of the time, which clearly makes no sense.

To solve the problem properly, avoiding any overlap among cases, you have to break the question down into non-overlapping situations, so you can correctly add the probabilities of each case. This is way too hard for the GMAT, but one way to do it is as follows:

case 1: all five bulbs fail, --the probability of this case is (0.15)^5
case 2: exactly four bulbs fail, which can happen in five ways, so the probability of this case is 5*(0.15)^4 * (0,85)
case 3: exactly three bulbs fail, and we do not have the sequence Fails, Works, Fails, Works, Fails (the only sequence with three failures where you don't have two consecutive failures) -- this can happen in 10 - 1 = 9 ways, so the probability of this case is 9 * (0.85)^2 * (0.15)^3
case 4: exactly two bulbs fail, and those two bulbs are consecutive, which can happen in 4 ways, so the probability of this case is 4 * (0.85)^3 (0.15)^2

Summing those probabilities with a good calculator, you get an answer slightly less than 0.08 (assuming I entered everything correctly). But there's no chance you'd ever need to answer a question like this on the GMAT.
If you are looking for online GMAT math tutoring, or if you are interested in buying my advanced Quant books and problem sets, please contact me at ianstewartgmat at gmail.com

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by Scott@TargetTestPrep » Thu Sep 05, 2019 5:05 am
BTGmoderatorDC wrote:A string of 5 different colored light bulbs is wired in such a way that if any two-consecutive light bulb fails, then entire string fails.
If for each individual light bulb, the probability of not failing during time period T is 0.85, what is the probability that the string of light bulbs will fail during the time period T?

Options
a) 0.0225
b) 0.09
c) 0.225
d) 0.25
e) 0.5
Since the probability of not failing is 0.85, the probability for failure is 1 - 0.85 = 0.15 for any given light bulb. Thus, if we take any two of these light bulbs, the probability that they both fail is (0.15)^2 = 9/400. Notice that in a string of 5 light bulbs, there are 4 pairs of light bulbs that are consecutive. Thus, the probability that a consecutive pair of light bulbs (and hence the whole system of light bulbs) fails is 4*9/400 = 9/100 = 0.09.

Answer: B

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scott@targettestprep.com

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