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# A string of 10 lightbulbs is wired in such a way that if any individua

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massi2884 wrote:
A string of 10 light bulbs is wired in such a way that if any individual light bulb fails, the entire string fails. If for each individual light bulb the probability of failing during time period T is 0.06, what is the probability that the string of light bulbs will fail during the time period T?

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

Cheers,
Brent
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Re: A string of 10 lightbulbs is wired in such a way that if any individua [#permalink]
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I understand why E is correct

but not understand why the result can not be 0.6. pls, explain using reasoning not using the quatitative.

why we can not add all possibilities?
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Re: A string of 10 lightbulbs is wired in such a way that if any individua [#permalink]
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thangvietnam wrote:
I understand why E is correct

but not understand why the result can not be 0.6. pls, explain using reasoning not using the quatitative.

why we can not add all possibilities?

SUGGESTED REASON:
Since P(A or B) = P(A) + P(B) – P(AnB)....if 2 events can happen at the same time P(AnB) is not 0
There exists a probability that 2 or more lightbulbs can fail at the same time
P(A or B or C) = P(A) + P(B) + P(C) – 2P(AnBnC) – [sum of exactly 2 groups members]

So for 10 items,
P(A or B or...or J) = P(A) + P(B)+...P(J) -...LOONG & COMPLEX calculations

I hope my reasoning is correct?
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Re: A string of 10 lightbulbs is wired in such a way that if any individua [#permalink]
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We are told that the entire wire will fail if one bulb fails. Since the probability of one bulb to fail is 0.6 then the probability of the entire wire to fail should be also 0.6.
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Dmba wrote:
We are told that the entire wire will fail if one bulb fails. Since the probability of one bulb to fail is 0.6 then the probability of the entire wire to fail should be also 0.6.

We are told that the string of light bulbs will fail if at least one light bulb fails. So, the favorable outcome is NOT one bulb to fail but the sum of the probabilities that one, two, three, ... or all 10 to fail OR which is the same, 1 minus the probability that neither fails.

Complete solution is here: a-string-of-10-light-bulbs-is-wired-in-such-a-way-that-if-131205.html

Hope it helps.
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Re: A string of 10 lightbulbs is wired in such a way that if any individua [#permalink]
Bunuel wrote:
Dmba wrote:
We are told that the entire wire will fail if one bulb fails. Since the probability of one bulb to fail is 0.6 then the probability of the entire wire to fail should be also 0.6.

We are told that the string of light bulbs will fail if at least one light bulb fails. So, the favorable outcome is NOT one bulb to fail but the sum of the probabilities that one, two, three, ... or all 10 to fail OR which is the same, 1 minus the probability that neither fails.

Complete solution is here: a-string-of-10-light-bulbs-is-wired-in-such-a-way-that-if-131205.html

Hope it helps.

Bunuel, can you solve this question with alternative method that you mentioned above -- the favorable outcome is NOT one bulb to fail but the sum of the probabilities that one, two, three, ... or all 10 to fail OR which is the same, [i]1 minus the probability that neither fails

Did not give the same result.
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Re: A string of 10 lightbulbs is wired in such a way that if any individua [#permalink]
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Ergenekon wrote:
Bunuel wrote:
Dmba wrote:
We are told that the entire wire will fail if one bulb fails. Since the probability of one bulb to fail is 0.6 then the probability of the entire wire to fail should be also 0.6.

We are told that the string of light bulbs will fail if at least one light bulb fails. So, the favorable outcome is NOT one bulb to fail but the sum of the probabilities that one, two, three, ... or all 10 to fail OR which is the same, 1 minus the probability that neither fails.

Complete solution is here: a-string-of-10-light-bulbs-is-wired-in-such-a-way-that-if-131205.html

Hope it helps.

Bunuel, can you solve this question with alternative method that you mentioned above -- the favorable outcome is NOT one bulb to fail but the sum of the probabilities that one, two, three, ... or all 10 to fail OR which is the same, [i]1 minus the probability that neither fails

Did not give the same result.

Hi Ergenekon,

Let me write the probability equation for event method:

P(String of lights failing) = P(1 or 2 or .....10 lights failing)

P(String of lights failing) = P(1 Fail & 9 Lit) + P(2 Fail & 8 Lit) +..........P( 10 Fail & 0 Lit)

P(String of lights failing) = $${10_C_1}$$ $$* 0.06^1*0.94^9 +$$ $${10_C_2}$$ $$* 0.06^2 *0.94^8 +$$.............$${10_C_{10}}$$ $$* 0.06^{10}*0.94^0$$

The important point here is to select the bulbs which would fail out of the total bulbs for each case. So, when we write the probability equation for 1 bulb failing, there can be 10 ways in which a bulb can fail.

Similarly, when you write the probability equation for the non-event method i.e.

P(String of lights failing) = 1 - P(All lights lit) $$= 1 -$$ $${10_C_{10}}$$ $$* 0.94^{10}$$

Both the probability equations will give you the same result. However you would notice that the non-event method is far more easier to calculate and comprehend than the event method.

For a probability question it is recommended to evaluate the number of cases in the event and the non-event method before proceeding with the solution.

Hope this helps

Regards
Harsh
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Re: A string of 10 lightbulbs is wired in such a way that if any individua [#permalink]
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For those not understanding this questing and wondering why you cannot do (6/100)^10

here is why

Let's make a simple example with using 2 lights and them each having a probability of failure of 1/10, and the chain fails if atleast 1 light fails.

What is the probability of failure?

Quick way: 1- P(success) --> 1 - (9/10)^2 = 19/100

Other Way

So why cant we do (1/10)^2 ... well we CAN we are just considering 1 case in which the strand fails. It also fails with light 1 success and light 2 fail

so we have

P(fail)*P(fail) --> 1/100
+
P(success)*P(fail) -> 9/100
+
P(fail)*P(success) -> 9/100

Together we get the probability of failure = 19/100

SO back to the 700 level question, the long way would be EXTREMELY cumbersome, so keep the 1-p(success) method in your mind ! The long way may only work for a 600 level question!
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It's important to realize that the question is asking for the probability that at least one light bulb will fail. When a question asks for the probability of "at least one", the easiest approach is to calculate the probability of "none" and take the complement of it.

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massi2884 wrote:
A string of 10 light bulbs is wired in such a way that if any individual light bulb fails, the entire string fails. If for each individual light bulb the probability of failing during time period T is 0.06, what is the probability that the string of light bulbs will fail during the time period T?

A. 0.06
B. (0.06)^10
C. 1 - (0.06)^10
D. (0.94)^10
E. 1 - (0.94)^10

We can use the formula:

P(at least one fails) = 1 - P(none of the 10 fail)

Since P(none of the 10 fail) = 0.94^10, then P(at least one fails) = 1 - 0.94^10.

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Re: A string of 10 lightbulbs is wired in such a way that if any individua [#permalink]
BrentGMATPrepNow wrote:
massi2884 wrote:
A string of 10 light bulbs is wired in such a way that if any individual light bulb fails, the entire string fails. If for each individual light bulb the probability of failing during time period T is 0.06, what is the probability that the string of light bulbs will fail during the time period T?

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

Cheers,
Brent

hey BrentGMATPrepNow why the answer choice is not B, Does this (0.06)^10 imply that all fall at the same time ?
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dave13 wrote:
BrentGMATPrepNow wrote:
massi2884 wrote:
A string of 10 light bulbs is wired in such a way that if any individual light bulb fails, the entire string fails. If for each individual light bulb the probability of failing during time period T is 0.06, what is the probability that the string of light bulbs will fail during the time period T?

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

Cheers,
Brent

hey BrentGMATPrepNow why the answer choice is not B, Does this (0.06)^10 imply that all fall at the same time ?

You're correct; (0.06)^10 is the probability that all 10 light bulbs fail at the same time.

That is P(all 10 light bulbs fail at the same time) = P(1st fails AND 2nd fails AND 3rd fails . . . . AND 9th fails AND 10th fails)
= P(1st fails) x P(2nd fails) x P(3rd fails) . . . . x P(9th fails) x P(10th fails)
= (0.06) x (0.06) x (0.06) x . . . (0.06) x (0.06) [b]
= (0.06)^10
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Re: A string of 10 lightbulbs is wired in such a way that if any individua [#permalink]
BrentGMATPrepNow wrote:
You're correct; (0.06)^10 is the probability that all 10 light bulbs fail at the same time.

That is P(all 10 light bulbs fail at the same time) = P(1st fails AND 2nd fails AND 3rd fails . . . . AND 9th fails AND 10th fails)
= P(1st fails) x P(2nd fails) x P(3rd fails) . . . . x P(9th fails) x P(10th fails)
= (0.06) x (0.06) x (0.06) x . . . (0.06) x (0.06)
= (0.06)^10

[b]BrentGMATPrepNow
thanks Brent, but doesnt option E stands for the same i mean that all 10 fall at the same time 1-94^10
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dave13 wrote:
BrentGMATPrepNow wrote:
You're correct; (0.06)^10 is the probability that all 10 light bulbs fail at the same time.

That is P(all 10 light bulbs fail at the same time) = P(1st fails AND 2nd fails AND 3rd fails . . . . AND 9th fails AND 10th fails)
= P(1st fails) x P(2nd fails) x P(3rd fails) . . . . x P(9th fails) x P(10th fails)
= (0.06) x (0.06) x (0.06) x . . . (0.06) x (0.06)
= (0.06)^10

[b]BrentGMATPrepNow
thanks Brent, but doesnt option E stands for the same i mean that all 10 fall at the same time 1-94^10

No, they're not the same.
We CAN say that (0.06)^10 = (1 - 0.94)^10, but we can't say that (0.06)^10 = 1 - 0.94^10
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Re: A string of 10 lightbulbs is wired in such a way that if any individua [#permalink]
Hello Bunuel,

I don't know where my reasoning is faltering. Can you please help me in locating the falter ?

My reasoning is that if any one of lightbulb fails, then the entire string will fails. And Probability of failing one light bulb is given the question stem. Therefore P(one lightbulb failing) = P (entire string failing).
So if it is 100 bulb string or 10 bulb string, the string will fail if any one of the bulb fails.
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Himanshuv0 wrote:
Hello Bunuel,

I don't know where my reasoning is faltering. Can you please help me in locating the falter ?

My reasoning is that if any one of lightbulb fails, then the entire string will fails. And Probability of failing one light bulb is given the question stem. Therefore P(one lightbulb failing) = P (entire string failing).
So if it is 100 bulb string or 10 bulb string, the string will fail if any one of the bulb fails.
Himanshuv0: you wrote "P(one lightbulb failing) = P (entire string failing)". This is false, because the probability of failure goes up when we are evaluating the entire string, instead of just one individual light bulb. The logic is that there are more opportunities for failure in the entire string. Similarly, our odds of winning a lottery go up if we buy more lottery tickets.

Or, let's illustrate with a dice game example:

Suppose we lose if we roll a "1" on any of the dice.

If we roll just 1 dice, our chances of losing are $$\frac{1}{6}$$.

However, what are the chances of losing if we roll 3 dice? Our chances of losing increase, because there are many possible losing combinations for the 3 dice:
ONN (One, Not One, Not One)
NON
NNO
OON
ONO
NOO
OOO

So, one way to solve would be to add all of those probabilities up.

However, a much easier way is using the formula: (1 - Probability of NOT a loss).

The probability of NOT a loss is: $$\frac{5}{6} * \frac{5}{6} * \frac{5}{6} = \frac{125}{216}$$

Therefore, the probability of a loss is $$1 - \frac{125}{216} = \frac{91}{216}$$
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