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Probability question

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Probability question [#permalink] New post 06 Sep 2005, 20:11
Could somebody please help me with this probability question--this was on the real GMAT...I'm not entirely sure of the numbers but the question is more or less the same.

There are three types of telephones:

telephone 1 - 65
telephone 2 - 75
telephone 3 - 110

total: 250

If two telephones are selected, what is the probability that both will be the same type of telephone?

(I'm reluctant to list the choices because I'm not sure if the numbers are correct)
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 [#permalink] New post 06 Sep 2005, 20:23
OK, here's my reasoning. I hope it's right:

The theory is that this is a pick without replacement question. The first pick is a free one; you have an 100% probability that the first telephone you pick will be one of the three types. The second one is the probability that, assuming you remove one of the first type, you'll pick one of the second type.

If the first one you pick is a Telephone 1, your chances of the second one being a Telephone 1 are 64/249.

If the first one you pick is a Telephone 2, your chances of the second one being a Telephone 2 are 74/249.

If the first one you pick is a Telephone 3, your chances of the second one being a Telephone 3 are 109/249.

But of course, each of those three scenarios has a different chance of happening. The first scenario has a 65/250 probability, the second scenario has a 75/250 probability and the third scenario has a 110/250 probability.

So the total probability = (65/250x64x249) + (75/250)(74/249) + (110/250)(109/249)

= 434/1245
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 [#permalink] New post 06 Sep 2005, 20:47
coffeeloverfreak wrote:
OK, here's my reasoning. I hope it's right:

The theory is that this is a pick without replacement question. The first pick is a free one; you have an 100% probability that the first telephone you pick will be one of the three types. The second one is the probability that, assuming you remove one of the first type, you'll pick one of the second type.

If the first one you pick is a Telephone 1, your chances of the second one being a Telephone 1 are 64/249.

If the first one you pick is a Telephone 2, your chances of the second one being a Telephone 2 are 74/249.

If the first one you pick is a Telephone 3, your chances of the second one being a Telephone 3 are 109/249.

But of course, each of those three scenarios has a different chance of happening. The first scenario has a 65/250 probability, the second scenario has a 75/250 probability and the third scenario has a 110/250 probability.

So the total probability = (65/250x64x249) + (75/250)(74/249) + (110/250)(109/249)

= 434/1245


Exactly what I did. Anyone else?
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 [#permalink] New post 06 Sep 2005, 20:48
here is how I view it, the result is the same as previous post:

favorable outcome:
65C2 + 75C2 + 110C2

all outcome:
250C2

the probablity is:

(65x64/2 + 75x74/2 + 110x109/2)/(250x249/2)

Last edited by qpoo on 06 Sep 2005, 22:00, edited 1 time in total.
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Re: Probability question [#permalink] New post 06 Sep 2005, 21:17
Thanks...that's what I thought. (after the fact)
If I remember correctly, the actual answer choices were clean decimals a) .15 b).20 c).30 d)0.35 e) .50

I'd be very impressed if anyone got the answer in less than two minutes.


jsun wrote:
Could somebody please help me with this probability question--this was on the real GMAT...I'm not entirely sure of the numbers but the question is more or less the same.

There are three types of telephones:

telephone 1 - 65
telephone 2 - 75
telephone 3 - 110

total: 250

If two telephones are selected, what is the probability that both will be the same type of telephone?

(I'm reluctant to list the choices because I'm not sure if the numbers are correct)
Re: Probability question   [#permalink] 06 Sep 2005, 21:17
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