string of 10 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 id 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
How do I approach these questions ?
Any pointer / book or example questions of similar kind would be very very helpful.
Thanks.
LightBulbs
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 theCodeToGMAT
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Answer would be {E}
Probability of Non Failure = 1  0.06 = 0.94
Probability of 10 bulbs not failing = (0.94)^10
So, probability of failure = 1  probability of non failure = 1  (0.94)^10
what is the OA?
Probability of Non Failure = 1  0.06 = 0.94
Probability of 10 bulbs not failing = (0.94)^10
So, probability of failure = 1  probability of non failure = 1  (0.94)^10
what is the OA?
R A H U L
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 Brent@GMATPrepNow
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ani781 wrote:string of 10 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 id 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
Aside: If P(bulb fails) = 0.06, then P(bulb doesn't fail) = 0.94
Okay, the entire string of lightbulbs will fail if 1 or more lightbulbs fail.
So, we want P(at least 1 lightbulb fails)
When it comes to probability questions involving "at least," it's best to try using the complement.
That is, P(Event A happening) = 1  P(Event A not happening)
P(at least 1 lightbulb fails) = 1  P(zero lightbulbs fail)
P(zero lightbulbs fail)
P(zero lightbulbs fail) = P(1st bulb doesn't fail AND 2nd bulb doesn't fail AND 3rd bulb doesn't fail AND . . . AND 9th bulb doesn't fail AND 10th bulb doesn't fail)
= P(1st bulb doesn't fail) x P(2nd bulb doesn't fail) x P(3rd bulb doesn't fail) x . . . x P(9th bulb doesn't fail) x P(10th bulb doesn't fail)
= (0.94) x (0.94) x (0.94) x . . . x(0.94) x (0.94)
= (0.96)^10
So, P(at least 1 lightbulb fails) = 1  P(zero lightbulbs fail)
= 1  (0.94)^10
= E
Cheers,
Brent
 ani781
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Thanks a lott Brent. Now this makes it very clear to me.
Is it possible for you to point me at some similar or almost in the same line of questions ? Or perhaps you can guide me to some more of these questions to further solidify my understanding ?
Thanks in advance.[/quote]
Is it possible for you to point me at some similar or almost in the same line of questions ? Or perhaps you can guide me to some more of these questions to further solidify my understanding ?
Thanks in advance.[/quote]
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 Brent@GMATPrepNow
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[/quote]ani781 wrote:Thanks a lott Brent. Now this makes it very clear to me.
Is it possible for you to point me at some similar or almost in the same line of questions ? Or perhaps you can guide me to some more of these questions to further solidify my understanding ?
Thanks in advance.
Sure thing.
Try these:
 https://www.beatthegmat.com/howtosolve ... 16924.html
 https://www.beatthegmat.com/probability ... 21264.html
 https://www.beatthegmat.com/probability ... 23109.html
 https://www.beatthegmat.com/birdt265208.html
 https://www.beatthegmat.com/probabilityt264072.html
Cheers,
Brent

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How do we solve it without using the complement method?
If the probability of one bulb failing is .06 is the probability of 10 bulb failing not= (.06)^10?
If the probability of one bulb failing is .06 is the probability of 10 bulb failing not= (.06)^10?
 DavidG@VeritasPrep
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(.06)^10 would be the probability of all 10 bulbs failing. But the entire string fails if at least one bulb fails. So there are going to be a lot of scenarios where this can happen.Gurpreet singh wrote:How do we solve it without using the complement method?
If the probability of one bulb failing is .06 is the probability of 10 bulb failing not= (.06)^10?
If 1 fails and 9 work, the string fails
If 2 fail and 8 work, the string fails
If 3 fail and 7 work, the string fails
etc.
It would take a very long time to calculate all of these individually. So you'd want to approach this the way Brent did, and solve for 1  P(no bulbs fail)

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