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# 10 light bulb

Author Message
TAGS:
Manager
Joined: 16 May 2011
Posts: 204
Concentration: Finance, Real Estate
GMAT Date: 12-27-2011
WE: Law (Law)
Followers: 1

Kudos [?]: 51 [0], given: 37

10 light bulb [#permalink]  06 Jul 2011, 11:48
00:00

Difficulty:

(N/A)

Question Stats:

43% (01:31) correct 57% (00:30) wrong based on 4 sessions
In a panel of 10 light bulbs, the odds of any light bulb to burn out are 0.06 at any given 10
minutes. Should a light bulb burn out, the whole panel would instantly turn off. What are
the odds that the panel would be turned off within the first 10 minutes after it was switched
on?
(A) 1-0.06^10
(B) 1-0.94^10
(C) (1-0.94)^10
(D) 0.06
(E) 1-10*0.06

Last edited by dimri10 on 06 Jul 2011, 12:30, edited 1 time in total.
Manager
Joined: 29 Jun 2011
Posts: 76
Followers: 3

Kudos [?]: 14 [1] , given: 47

Re: 10 light bulb [#permalink]  06 Jul 2011, 11:59
1
KUDOS
IMO B...but is B [1 - (0.94^10)] because thats what I am getting.

solution : consider in 10 minutes space
1 light bulb can go off P = 0.06
so, P that the lightbulb does not go off = 1-0.06 =.94
now we have 10 lightbulbs. Their functioning properly forms a mutually exclusive set of events.
therefore, P that all of them function properly = .94 * .94 (10 times) = .94^10
now, for the whole system to trip even 1 faulty bulb is enough.
therefore, we need P(at least one trips)
which is P(at least one) = 1 - P(none trip) = 1 - P(all function properly) = 1 - 0.94^10
Ans B
_________________

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How charged with punishments the scroll,
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Manager
Joined: 16 May 2011
Posts: 204
Concentration: Finance, Real Estate
GMAT Date: 12-27-2011
WE: Law (Law)
Followers: 1

Kudos [?]: 51 [0], given: 37

Re: 10 light bulb [#permalink]  06 Jul 2011, 12:30
fivedaysleft wrote:
IMO B...but is B [1 - (0.94^10)] because thats what I am getting.

solution : consider in 10 minutes space
1 light bulb can go off P = 0.06
so, P that the lightbulb does not go off = 1-0.06 =.94
now we have 10 lightbulbs. Their functioning properly forms a mutually exclusive set of events.
therefore, P that all of them function properly = .94 * .94 (10 times) = .94^10
now, for the whole system to trip even 1 faulty bulb is enough.
therefore, we need P(at least one trips)
which is P(at least one) = 1 - P(none trip) = 1 - P(all function properly) = 1 - 0.94^10
Ans B

EDITED. THANK'S FOR NOTICING
Manager
Joined: 14 Feb 2011
Posts: 69
Followers: 1

Kudos [?]: 1 [0], given: 2

Re: 10 light bulb [#permalink]  08 Aug 2011, 11:52
dimri10 wrote:
In a panel of 10 light bulbs, the odds of any light bulb to burn out are 0.06 at any given 10
minutes. Should a light bulb burn out, the whole panel would instantly turn off. What are
the odds that the panel would be turned off within the first 10 minutes after it was switched
on?
(A) 1-0.06^10
(B) 1-0.94^10
(C) (1-0.94)^10
(D) 0.06
(E) 1-10*0.06

ans B.

Prob that light will not turned off = 0,94^10. Hence prob (turn off) = 1 - prob (not turn off).
Re: 10 light bulb   [#permalink] 08 Aug 2011, 11:52
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