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A string of 10 lightbulbs is wired in such a way that if any
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07 Sep 2020, 04:45
07 Sep 2020, 04:45
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Question Stats:
51% (01:05) correct
48% (01:04) wrong based on 192 sessions
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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 is 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. \(1 - (0.94)^{10}\)
A. \(0.06\)
B. \((0.06)^{10}\)
C. \(1 - (0.06)^{10}\)
D. \((0.94)^{10}\)
E. \(1 - (0.94)^{10}\)
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Re: A string of 10 lightbulbs is wired in such a way that if any
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07 Sep 2020, 06:00
07 Sep 2020, 06:00
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Carcass wrote:
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 is 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. \(1 - (0.94)^{10}\)
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
Answer: E
Cheers,
Brent
General Discussion
Re: A string of 10 lightbulbs is wired in such a way that if any
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07 Sep 2020, 04:46
07 Sep 2020, 04:46
Expert Reply
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