A 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
Ans is E
I think the answer is C, as
The probability that the string of lightbulbs will fail during time period T= 1 – The probability that the string of lightbulbs will not fail during time period T
The probability that the string of lightbulbs will not fail during time period T = .06^10
So the probability that the string of lightbulbs will fail during time period T= 1 – .06^10
prob of light string fails = 1 – P(light strings not fail)
P(light string not fail) = 0.94
as there r 10 bulbs so
required prob is 1-(0.94)^10
since the bulbs are connected serially, all is required that one bulb fails for the entire string to fail. so it is .06.
A.
E…
Prob. of Light bulb not failing = 0.96
Prob. string of light bulbs doesn’t fail = 0.96 x 0.96 x… x 0.96 (10 times) [since 1st lb doesn't fair AND 2nd lb doesnt fail, etc.]
Therefore, prob. that it WILL fail: 1 – (0.96)^10
Since the string of lb will fail even if one of the lb fails,
P(String of lb fails) = (Number of ways one of the lb out of 10 lb fails)*P(lb fails)
= C(10,1)*0.06
= 10*0.06 = 0.6
What am I doing wrong?
( * means multiplied)
@hari
The string would fail even if more than one bulb fails.
Hari … the probability that one bulb fails is …
= P(one bulb fails ) * p( 9 bulbs do not fail)
then you also have to consider the case when 2 , 3, 4 … bulbs fail.
That is the reason it is easy to calculate the case that when non fail and subtract it from 1.
There is nothing wrong with Hari’s solution. It’s just that hari’s answer is not there in the options. So…we need to take a different approach to the question. .06 is the probability of any bulb failing. hence, 0.94 is the probability of not failing. we know that the string will fail if any of the bulb fails. other way of looking at this is that the string will fail if atleast one bulb fails…that is… 1-Probability of no bulb failing. Hence the answer is
1-(0.94)^10.
Let p=no. of defective bulbs,q=no. of non-defective bulbs.
Since, it has been given that if any one bulb fails the entire string will go off.Therefore,string can go off when 1,2,3,4,………………10 nos. of bulb will go off.
Given q=0.94 & p=0.06.
Therefore by binomial theoram,the probability for the strings to go off is 10C(0.06)^1(0.94)^9……………………………..10C10(0.06)^10 (0.94)^0=1-10C(0.06)^0(0.94)^10=1-(0.94)^10.
Hence, E
b
A
The answer is E. As rightly discussed P(fail) = 1 – P(does not fail)
P(1 bulb does not fail) = 1-0.6 = 0.94. like wise there are 10 bulbs and each one of them should not fail. so total probability of not failing is 0.94^10.
so P(fail) = 1 – 0.94^10
E is the right answer
E
One Bulb fails X 9 bulbs dont fall + Two bulb fail X 8 bulbs dont fall + ….+ all bulbs fall
=> c(n,1)0.6 X c(n,9)0.94 + c(n,2)0.6 X c(n,8)0.94 +…..+ c(n,10)0.6
= (0.6 + 0.94) ** 10 – 0.94 ** 10
= 1 – (0.94) ** 10
p=bulb does not=0.94 fail q=bulb fail=0.06
now, Total prob(none f the bulb fails)=0.94*0.94*0.94….. for 10 bulbs=0.94^10
now prob(atleast one bulb fails)=1-prob(none f the bulb fails)=1-0.94^10
Ans: E
Confident of E
e is correct.
Ans A
Since the bulbs are connected in series, therefore any of the bulb’s failure will result in the failure of the string. it is irrelevant whether one or more than one bulbs have failed.
hm..what my logic is the minimum possibility of failure of string is when all the bulbs simultaneously fail (by mentioned probability). Thus it’ll bcome the string’s probability. i. e. 0.06
The max. Probability would be in the case when each bulb fails separately with no intersection of other bulb’s failure. thus, probability will bcome 0.06*10= 0.6
so how to combine these to ans..(because ans must lie btwn 0.06 to 0.6)
hope i m clear n wr8 as well..
Awesome question!! E. kms explained it well!
I say it is B, the probability of one failing is 0.06 but there are 10 bulbs, so if any of the 10 fail the whole string will fail which is what we are looking for. Therefore you have to look at each bulb separately. Prob of one 0.06, prob of the second 0.06, prob of the third 0.06, etc. These all have to be multiplied together to get the probablity of any of the bulbs failing which would cause the string to fail.
Thanks.. hope im right
I will go with E
But logic equally supports choice A.
I’m confused??
Gmat Team ,Need your help
C
E for me as well…
answer should be (o.o6*10) which is 0.6 which means E
E…
Should be A
E
Can we have the official answer please???
Confused in A & B.
E …
the answer is A hat because it is said that if any individual lightbulb fails, the entire string fails.so the probability of failing of just one lighbulb is the probability of failing of the entire sring.