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A string of 10 light bulbs is wired in such a way that if [#permalink]
23 Apr 2012, 13:43

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A

B

C

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E

Difficulty:

45% (medium)

Question Stats:

58% (01:42) correct
42% (00:39) wrong based on 227 sessions

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

I know it's not among the answer choices, but could you please tell me what's wrong with thinking that the result should be 0.06 + 0.06 + 0.06 + 0.06 + 0.06 + 0.06 + 0.06 + 0.06 + 0.06 + 0.06 = 10*0.06 = 0.6 ?

Re: A string of 10 light bulbs is wired in such a way that if [#permalink]
23 Apr 2012, 20:37

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Quote:

I know it's not among the answer choices, but could you please tell me what's wrong with thinking that the result should be 0.06 + 0.06 + 0.06 + 0.06 + 0.06 + 0.06 + 0.06 + 0.06 + 0.06 + 0.06 = 10*0.06 = 0.6 ?

It only takes one failure in any of the 10 slots to create the failure. For example if the second bulb fails, the whole string fails. If the 9th bulb fails, the string will fail too. Therefore the one and only chance for the strings to stay lit would be 10 consecutive non-failures.

(.94)^10 = a lit string. Now it becomes 1-(.94)^10 since you're looking for the probability that the string of lights will fail.

Re: A string of 10 light bulbs is wired in such a way that if [#permalink]
24 Apr 2012, 12:12

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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

I know it's not among the answer choices, but could you please tell me what's wrong with thinking that the result should be 0.06 + 0.06 + 0.06 + 0.06 + 0.06 + 0.06 + 0.06 + 0.06 + 0.06 + 0.06 = 10*0.06 = 0.6 ?

The string of light bulbs will fail if at least one light bulb fails. So, let's find the probability of the opposite event and subtract that value from 1.

The opposite event is when none of the 10 light bulbs fails, since the probability of each light bulb not to fail is 1-0.06=0.94 the the probability that none of the 10 light bulbs fails is 0.94^10.

Hence, the probability that at least one light bulb fails is 1-0.94^10.

Answer: E.

Now, you should have spotted that your reasoning was not right because of one simple thing, consider the case when we have 100 light bulbs instead of 10, then according to your logic the probability that the string of light bulbs will fail would be 100*0.06=6, which is not possible since the probability of an event cannot be more than 1 (100%).

A string of 10 lightbulbs is wired in such a way that if any [#permalink]
16 May 2012, 02:41

A string of 10 lightbulbs is wired in such a way that if any individal lightbulb fails, the entire string fails. If for each individual lightbulb the probability of failing during the period T is 0.06, what is the probability that the string of lightbulbs will fail during the period T?

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

I still believe it is B. Why exactly is it E? I understand that E is 1- the probability of NOT failing. But I would assume the B is just the probability of it failing.

Re: A string of 10 lightbulbs is wired in such a way that if any [#permalink]
16 May 2012, 02:47

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Expert's post

Merging similar topics.

alexpavlos wrote:

A string of 10 lightbulbs is wired in such a way that if any individal lightbulb fails, the entire string fails. If for each individual lightbulb the probability of failing during the period T is 0.06, what is the probability that the string of lightbulbs will fail during the period T?

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

I still believe it is B. Why exactly is it E? I understand that E is 1- the probability of NOT failing. But I would assume the B is just the probability of it failing.

We are told that the string of light bulbs will fail if at least one light bulb fails. (0.06)^10 is the probability that all 10 lightbulbs will fail.

Re: A string of 10 lightbulbs is wired in such a way that if any [#permalink]
16 May 2012, 03:45

Bunuel wrote:

Merging similar topics.

alexpavlos wrote:

A string of 10 lightbulbs is wired in such a way that if any individal lightbulb fails, the entire string fails. If for each individual lightbulb the probability of failing during the period T is 0.06, what is the probability that the string of lightbulbs will fail during the period T?

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

I still believe it is B. Why exactly is it E? I understand that E is 1- the probability of NOT failing. But I would assume the B is just the probability of it failing.

We are told that the string of light bulbs will fail if at least one light bulb fails. (0.06)^10 is the probability that all 10 lightbulbs will fail.

Re: A string of 10 light bulbs is wired in such a way that if [#permalink]
17 Jan 2013, 01:20

1

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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? _________________

KUDOS me if you feel my contribution has helped you.

Re: A string of 10 light bulbs is wired in such a way that if [#permalink]
16 Jan 2014, 22:28

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

I know it's not among the answer choices, but could you please tell me what's wrong with thinking that the result should be 0.06 + 0.06 + 0.06 + 0.06 + 0.06 + 0.06 + 0.06 + 0.06 + 0.06 + 0.06 = 10*0.06 = 0.6 ?

Probability of Failing = 1 - probability of success Probability of failing = 1 - (1 - 0.06)^10 = 1 - (0.94)^10 _________________

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Re: A string of 10 light bulbs is wired in such a way that if [#permalink]
17 Jan 2014, 05:37

Can please someone explain why answer A is wrong? 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.

Re: A string of 10 light bulbs is wired in such a way that if [#permalink]
17 Jan 2014, 05:46

Expert's post

Dmba wrote:

Can please someone explain why answer A is wrong? 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.

Re: A string of 10 light bulbs is wired in such a way that if [#permalink]
31 Jan 2015, 21:30

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Re: A string of 10 light bulbs is wired in such a way that if [#permalink]
11 May 2015, 23:41

Bunuel wrote:

Dmba wrote:

Can please someone explain why answer A is wrong? 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.

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. _________________

If my post was helpful, press Kudos. If not, then just press Kudos !!!

Re: A string of 10 light bulbs is wired in such a way that if [#permalink]
12 May 2015, 01:41

1

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Expert's post

Ergenekon wrote:

Bunuel wrote:

Dmba wrote:

Can please someone explain why answer A is wrong? 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.

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

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.

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.

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