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

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EgmatQuantExpert
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A string of 5 different colored light bulbs is wired in such a way tha [#permalink] New post  Updated on: 27 Apr 2018, 05:04
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3 deadly mistakes you must avoid in Probability - Exercise Question #1


1- 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



To solve question 2: Question 2

To read the article: 3 deadly mistakes you must avoid in Probability
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Originally posted by EgmatQuantExpert on 26 Apr 2018, 04:29.
Last edited by EgmatQuantExpert on 27 Apr 2018, 05:04, edited 2 times in total.
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A string of 5 different colored light bulbs is wired in such a way tha [#permalink] New post  Updated on: 29 Apr 2018, 01:31
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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
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Originally posted by EgmatQuantExpert on 26 Apr 2018, 04:30.
Last edited by EgmatQuantExpert on 29 Apr 2018, 01:31, edited 1 time in total.
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Re: A string of 5 different colored light bulbs is wired in such a way tha [#permalink] New post  26 Apr 2018, 05:23
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EgmatQuantExpert wrote:
3 deadly mistakes you must avoid in Probability - Exercise Question #1


1- 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


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
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Re: A string of 5 different colored light bulbs is wired in such a way tha [#permalink] New post  29 Apr 2018, 01:33
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Hey everyone,

The official answer to the question has been posted.

Regards,
Ashutosh
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Re: A string of 5 different colored light bulbs is wired in such a way tha [#permalink] New post  05 Jun 2020, 07:32
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?
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robinbjoern
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Re: A string of 5 different colored light bulbs is wired in such a way tha [#permalink] New post  02 Jul 2020, 05:52
EgmatQuantExpert wrote:

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



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?
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josenetofaria
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Re: A string of 5 different colored light bulbs is wired in such a way tha [#permalink] New post  15 Aug 2020, 06:55
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robinbjoern wrote:
EgmatQuantExpert wrote:

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



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?


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.)
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harmanvirdi90
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A string of 5 different colored light bulbs is wired in such a way tha [#permalink] New post  11 Sep 2021, 03:18
EgmatQuantExpert wrote:

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


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 ?
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