The probability of a vehicle failing, PLEASE!

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The probability of a vehicle failing, PLEASE!

by navai » Thu Aug 26, 2010 7:23 am
Hello, guys.
I have a test and can't solve a few questions. Could you give a hand of help, please?

First question.
The probability of a machine breaking down on any single day is 0.05. What is the probability that there will be 2 or more breakdowns over a 14 day period?

choice the right answer :
0.415
0.229
0.771
0.153
0.842
Thank you
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by akhpad » Thu Aug 26, 2010 8:40 am
Is it 0.153?

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by scorpionz » Thu Aug 26, 2010 11:12 am
Yes..I too came up with an answer of 0.153..But there's no way in hell that I could've done it without a calculator!

What we are essentially looking out for is...

Probability of 2 or more breakdowns in a 14 day period
=1-[(probability of only 1 breakdown in the 14 day period) + (probability of no breakdown in the 14 day period)]
=1-[(0.05)*(0.95^13)*(14) + (0.95^14)]
=0.153

I can't even think of a method to make an educated guess if such a problem were to come in the actual test!!

Keeping my fingers crossed that I'm not so unlucky!! :)

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by Ian Stewart » Thu Aug 26, 2010 12:55 pm
scorpionz wrote:Yes..I too came up with an answer of 0.153..But there's no way in hell that I could've done it without a calculator!

What we are essentially looking out for is...

Probability of 2 or more breakdowns in a 14 day period
=1-[(probability of only 1 breakdown in the 14 day period) + (probability of no breakdown in the 14 day period)]
=1-[(0.05)*(0.95^13)*(14) + (0.95^14)]
=0.153

I can't even think of a method to make an educated guess if such a problem were to come in the actual test!!
That solution is perfect.

The GMAT is not testing if you are a human Excel spreadsheet, so you will never be asked to perform any absurdly difficult calculations on the test. The question in this thread tests something known as 'binomial probability'. While I've seen one or two real GMAT questions that test this topic, it very rarely appears, and when it does, if the numbers are at all awkward, the answer choices are left as 'formulas'. That is, they are testing if you know what you'd need to type into your calculator to get the answer - they aren't testing if you can calculate the answer yourself. If this question were a real GMAT question (it definitely is *not* one as written) the answer would be written in the form you see in the second last line of your solution, above the line '=0.153'. That said, the numbers here seem too inelegant to be likely on the GMAT; the GMAT normally tests difficult concepts using simple numbers.
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by BastiG » Thu Aug 26, 2010 3:04 pm
scorpionz wrote:Yes..I too came up with an answer of 0.153..But there's no way in hell that I could've done it without a calculator!

What we are essentially looking out for is...

Probability of 2 or more breakdowns in a 14 day period
=1-[(probability of only 1 breakdown in the 14 day period) + (probability of no breakdown in the 14 day period)]
=1-[(0.05)*(0.95^13)*(14) + (0.95^14)]
=0.153

I can't even think of a method to make an educated guess if such a problem were to come in the actual test!!

Keeping my fingers crossed that I'm not so unlucky!! :)
Can someone pls explain why (0.05)*(0.95^13)) is multiplied with (14).

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by Ian Stewart » Thu Aug 26, 2010 3:42 pm
BastiG wrote:

Can someone pls explain why (0.05)*(0.95^13)) is multiplied with (14).
14*(0.05)*(0.95^13) represents the probability that exactly one of the machines breaks down. You could first ask - what's the probability the first machine breaks down, and the other thirteen machines work? This would be equal to

(0.05)*(0.95)*(0.95)*...*(0.95) = (0.05)*(0.95^13)

Now, if we find the probability the second machine breaks down and the others work, this will be equal to

(0.95)*(0.05)*(0.95)*(0.95)*...*(0.95) = (0.05)*(0.95^13)

Similarly, the probability the third machine will be the only one that breaks down will be (0.05)*(0.95^13), and so on. We'll get 14 individual probabilities like this. To find the probability that any one machine breaks down, we'd add the probability that only the first breaks down, the probability that only the second breaks down, the probability that only the third breaks down, etc, so we'd get 14*(0.05)(0.95^13).

That's the long explanation, just to explain why this faster method works: you can first find the probability that one machine fails in some specific sequence: (0.05)(0.95^13). You can then multiply this by the number of sequences in which exactly one machine fails: 14.

I'd add that this type of probability question is only very rarely asked on the GMAT, and it is *far* more important to become comfortable with the more basic concepts of probability. If you are comfortable with the more elementary types of probability questions, you might, however, try to apply the above to a question like the following, which has numbers more like what you'd see on the GMAT:

In a certain city, the probability of rain on any given day is 0.7. What is the probability it rains on exactly one day in a certain five day period?
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by BastiG » Fri Aug 27, 2010 6:20 pm
Ian Stewart wrote: In a certain city, the probability of rain on any given day is 0.7. What is the probability it rains on exactly one day in a certain five day period?
So i guess the answer should be:

(0,7)*(0,3)^4*5
=7*0,081*5
=0,567*5
=2,835

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by Ian Stewart » Fri Aug 27, 2010 7:42 pm
BastiG wrote:
Ian Stewart wrote: In a certain city, the probability of rain on any given day is 0.7. What is the probability it rains on exactly one day in a certain five day period?
So i guess the answer should be:

(0,7)*(0,3)^4*5
=7*0,081*5
=0,567*5
=2,835
Yes, exactly right (though the answer you give in the final line is a percentage; as a probability, it would be 0.02835). :)

On the GMAT, the answer might well be left in the form 5*(0.7)*(0.3)^4 in the answer choices, since it isn't all that interesting to actually carry out the calculation, and the GMAT rarely requires you to perform lengthy computations. Still, with these numbers, it's not all that bad to do.
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by diebeatsthegmat » Sun Sep 19, 2010 12:29 pm
Ian Stewart wrote:
scorpionz wrote:Yes..I too came up with an answer of 0.153..But there's no way in hell that I could've done it without a calculator!

What we are essentially looking out for is...

Probability of 2 or more breakdowns in a 14 day period
=1-[(probability of only 1 breakdown in the 14 day period) + (probability of no breakdown in the 14 day period)]
=1-[(0.05)*(0.95^13)*(14) + (0.95^14)]
=0.153

I can't even think of a method to make an educated guess if such a problem were to come in the actual test!!
That solution is perfect.

The GMAT is not testing if you are a human Excel spreadsheet, so you will never be asked to perform any absurdly difficult calculations on the test. The question in this thread tests something known as 'binomial probability'. While I've seen one or two real GMAT questions that test this topic, it very rarely appears, and when it does, if the numbers are at all awkward, the answer choices are left as 'formulas'. That is, they are testing if you know what you'd need to type into your calculator to get the answer - they aren't testing if you can calculate the answer yourself. If this question were a real GMAT question (it definitely is *not* one as written) the answer would be written in the form you see in the second last line of your solution, above the line '=0.153'. That said, the numbers here seem too inelegant to be likely on the GMAT; the GMAT normally tests difficult concepts using simple numbers.
this means we can ignore this problem