A string of 5 different colored light bulbs is wired in such a way

Question

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?

A. 0.0225
B. 0.09
C. 0.225
D. 0.25
E. 0.5

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EgmatQuantExpert · Accepted Answer

Solution

Given:

• There is a string of 5 different colored light bulbs.
• The string fails if any two-consecutive light bulb fails.
• The probability of not failing of each bulb during time period T= 0.85

To find:

• We need to find the probability that the string of light bulbs will fail during the time period T.

Approach and Working

• Since the string fails only when two consecutive bulbs fail, 
 o Hence, P (String will fail) = P (one bulb fails) * P (another consecutive bulb fails) * ways in which two consecutive light bulbs can be chosen

• We know P (Not failing of a bulb) = 0.85
 o Thus, P’ (failing of bulb) = 0.15

Total ways in which two light bulbs can fail: 
 • The total ways to select to select 2 consecutive light bulbs in a string of 5 bulbs = 4.

Thus, P (String will fail) = 0.15* 0.15*4=0.0225*4= 0.09

Hence, the correct answer is option B.

Answer: B

viv007 · Answer

hi...

we have to find the probability of fail...
and we have given that if two consecutive bulbs fail then only the complete string will fail.
so..
given prob. of not fail = .85 
so 
prob. of fail = .15

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

prob of fail = .15*.15 * 4
 = .09 
ans B

Kritisood · Answer

if we consider probability that the string of light bulbs will not fail,
so i.e 0.85^5
so probability will fail is 1- 0.85^5 => 0.5

I know im not taking into consideration the prob of consecutive failing but still not able to logically understand why this is wrong.

chetan2u   Bunuel  could you pls help?

robinbjoern · Answer

why don't we need to add 0.85^3 at the end?

is it because the question is not asking about the concrete probability of 2 light bulbs failing?

josenetofaria · Answer

Actually, it doesn't matter if the other bulbs will fail or not.

Pay attention to the criteria of failure: " It is ENOUGH that two consecutive bulbs fail for the failure of the whole string "

So, to illustrate this, repair that we have:

F = Fails
W = Works
 
FFWWW 
FFWWF

In any of these cases, we have a general failure of the string, because the criteria of " Two consecutive have to fail " was reached.
If we were to illustrate the probability of both of the events above occur, we would have equal probabilities:

P(failure) = 0,15*0,15. (Again, it won't matter what is going to happen with the other bulbs, since two consecutives of them have failed.)

harmanvirdi90 · Answer

EgmatQuantExpert

Let's suppose A' denotes probability of failing of bulb and A denotes probability of not failing

Conditions of failure of string
1) A'A' i.e first bulb fails and the second bulb fails : P(1): (.15)*(.15)
2) AA'A': First is OK and second and third bulb fails : P(2) : (.85)*(.15)*(.15)
3) AAA'A': P(3): (.85)^2*(.15)^2
4) AAAA'A' : P(4): (.85)^3*(.15)^2

Combine Probability : P(1)+P(2)+P(3)+P(4) = 0.083

Please tell me where I did mistake ?

tgmat24 · Answer

It is true that it is sufficient for the string to fail if any 2 consecutive bulbs stop functioning, however, it’s not necessary that if the entire string has stopped functioning it would mean that only 2 consecutive bulbs have stopped during time "T", it could also mean that 3 consecutive, 4 consecutive or all the bulbs have stopped functioning. Hence, to calculate the probability, we must take all those cases into account as well.

egmat please clarify

egmat please clarify

vijayakrishnan · Answer

Agreed. Bad question

Ddefines · Answer

The no. of ways to choose 2 bulbs out of a string of 5 bulbs i.e., 5C2 is not 4, it is 10.
I don't know why it's mentioned as 4 in the above solutions.
Can someone explain?

tgmat24 · Answer

We have to choose consecutive pairs and not just any 2 bulbs out of 5.

Vitthal19 · Answer

in this answer we have taken all the cases when light fails when 2 consecutive fails, but it will also fail when 3 or 4 or 5 consecutive won't work so aren't we supposed to take them into account while solving this question?

onlymalapink · Answer

we dont have to take other fails in account because the whole system will fail before any other consecutive failure has a chance to occur

skansal1999 · Answer

Hey, can somebody provide a detailed solution for this one. Having a hard time solving this one.

A string of 5 different colored light bulbs is wired in such a way

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A string of 5 different colored light bulbs is wired in such a way

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