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LightBulbs

by ani781 » Thu Sep 26, 2013 4:12 am
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.
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by theCodeToGMAT » Thu Sep 26, 2013 4:56 am
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?
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by ani781 » Thu Sep 26, 2013 5:25 am
You are really nice at these. Thanks; your answer is indeed the OA.
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by Brent@GMATPrepNow » Thu Sep 26, 2013 5:28 am
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
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by ani781 » Thu Sep 26, 2013 5:37 am
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]
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by Brent@GMATPrepNow » Thu Sep 26, 2013 6:00 am
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.
[/quote]

Sure thing.
Try these:
- https://www.beatthegmat.com/how-to-solve ... 16924.html
- https://www.beatthegmat.com/probability- ... 21264.html
- https://www.beatthegmat.com/probability- ... 23109.html
- https://www.beatthegmat.com/bird-t265208.html
- https://www.beatthegmat.com/probability-t264072.html

Cheers,
Brent
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by Gurpreet singh » Tue Jun 28, 2016 5:58 am
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?
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by DavidG@VeritasPrep » Tue Jun 28, 2016 7:48 am
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?
(.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.

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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by Gurpreet singh » Tue Jun 28, 2016 4:29 pm
Thank you.
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