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A buyer buys 3 different items out of the newly introduced 10 differen

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A buyer buys 3 different items out of the newly introduced 10 differen [#permalink]

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New post 27 May 2010, 11:34
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A buyer buys 3 different items out of the newly introduced 10 different items. If two items were to be selected at random, what is the probability that the buyer does not have both the chosen items?

If possible, could people walk through steps and logic taken because I got an answer completely different than what was given.

Thanks

[Reveal] Spoiler:
Answer given was 7/15 with the logic of 7C2/10C2. IMO, that logic answers the question of the buyer not having either of the items. Am I wrong?
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Re: A buyer buys 3 different items out of the newly introduced 10 differen [#permalink]

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New post 27 May 2010, 11:54
The question is worded a little poorly - when it says "both" it actually means that the buyer did not buy either. Both could also indicate that it is possible the buyer had just 1 of them, which complicates the math as I'm sure you must have realized. Don't worry - in the GMAT it will be clearer.

The buyer has bought 3 out of the 10 possible items.

The total number of possibilities of choosing 2 out of the 10 items is \(10C2 = 45\).

The total number of items which the buyer did not buy \(= 10 -3 = 7\)

Therefore the number of possibilities of the 2 items being chosen from these 7 = \(7C2 = 21\)

Therefore probability that the 2 items chosen are NOT the 3 that the buyer bought is:
\(\frac{21}{45} = \frac{7}{15}\)

Saying that the buyer bought 3 items which were NOT these 2 items is equivalent to saying that we chose 2 items neither of which were any of the 3 that the buyer bought.

Here's the proof:

Total number of ways the buyer can choose 3 items out of 10 = \(10C3 = 120\).

Number of items which were NOT chosen \(= 10 -2 = 8\)

Therefore the number of possibilities of the buyer buying the 3 items from these 8 = \(8C3 = 56\).

The probability = \(\frac{56}{120} = \frac{7}{15}\)

Same answer ;)
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Re: A buyer buys 3 different items out of the newly introduced 10 differen [#permalink]

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New post 27 May 2010, 12:01
A quick way to do a problem like this without having to think about "X choose Y" is the following:

The probability you're looking for is:

He doesn't select one of the items he already has with the first choice

AND (multiply)

He doesn't select one of the items he already has with the second choice

Probability he doesn't select something he already has with first choice = 7 / 10

" " second choice = 6/9

=> 7/10 * 6/9

= 42/90 = 7/15
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Re: A buyer buys 3 different items out of the newly introduced 10 differen [#permalink]

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New post 27 May 2010, 12:48
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nphilli1 wrote:
A buyer buys 3 different items out of the newly introduced 10 different items. If two items were to be selected at random, what is the probability that the buyer does not have both the chosen items?

If possible, could people walk through steps and logic taken because I got an answer completely different than what was given.

Thanks

[Reveal] Spoiler:
Answer given was 7/15 with the logic of 7C2/10C2. IMO, that logic answers the question of the buyer not having either of the items. Am I wrong?


The answer 7/15 implies that the question means that buyer has none of the items selected - \(\frac{C^2_7}{C^2_{10}}=\frac{7}{15}\).

But "does not have both the chosen items" in my opinion should mean that buyer has none of the items selected OR has only one of the items selected BUT not both of them.

In this case: probability = 1 - probability of having both = \(1-\frac{C^2_3}{C^2_{10}}=\frac{14}{15}\).

Anyway not a good question.
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Re: A buyer buys 3 different items out of the newly introduced 10 differen [#permalink]

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New post 22 Apr 2011, 14:46
The way I thought about it was, what is the chance that you pick 3 that are not part of the 2 chosen.

Therefore:

(8/10) * (7/9) * (6/8) = 346/720 = 7/15

Everytime you choose one, you avoid the 2 "randomly" selected ones.

When you choose out of 10, there are 8; when you choose the next one (with one removed), you choose out of 9, with 7 choices; etc.
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Re: A buyer buys 3 different items out of the newly introduced 10 differen [#permalink]

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New post 26 Oct 2011, 12:08
Hi,

Someone pls correct me here...
P(choosing neither) = 1 - P(choosing either or both)

therefore,
P(ch. neither) = 1 - [(2C1.8C2 + 2C1.8C2 + 2C2.8C1)/10C3]
= 1 - (128/120).

Pls help.

Thanks
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Re: A buyer buys 3 different items out of the newly introduced 10 differen [#permalink]

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New post 26 Oct 2011, 22:38
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raghupara wrote:
Hi,

Someone pls correct me here...
P(choosing neither) = 1 - P(choosing either or both)

therefore,
P(ch. neither) = 1 - [(2C1.8C2 + 2C1.8C2 + 2C2.8C1)/10C3]
= 1 - (128/120).

Pls help.

Thanks


This is what the question means: (there is a debate on the intent due to the OA but this is what the question means to me)
Out of 10 items, a man buys 3. Now you have to pick any two items out of the original 10. What is the probability that both your items were not among those selected by the man too? (It is possible that one of your items match but both should not.)

P(both your items were not selected by man) = 1 - P(both items were selected by man) = 1 - 3C2/10C2 = 1 - 3/45 = 14/15

Intent as depicted by the OA:
What is the probability that neither of your items was among those selected by the man?

P(choosing neither of the man's items) = 7C2/10C2 = 21/45 = 7/15
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Re: A buyer buys 3 different items out of the newly introduced 10 differen [#permalink]

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New post 27 Oct 2011, 00:59
VeritasPrepKarishma wrote:
raghupara wrote:
Hi,

Someone pls correct me here...
P(choosing neither) = 1 - P(choosing either or both)

therefore,
P(ch. neither) = 1 - [(2C1.8C2 + 2C1.8C2 + 2C2.8C1)/10C3]
= 1 - (128/120).

Pls help.

Thanks


This is what the question means: (there is a debate on the intent due to the OA but this is what the question means to me)
Out of 10 items, a man buys 3. Now you have to pick any two items out of the original 10. What is the probability that both your items were not among those selected by the man too? (It is possible that one of your items match but both should not.)

P(both your items were not selected by man) = 1 - P(both items were selected by man) = 1 - 3C2/10C2 = 1 - 3/45 = 14/15

Intent as depicted by the OA:
What is the probability that neither of your items was among those selected by the man?

P(choosing neither of the man's items) = 7C2/10C2 = 21/45 = 7/15



Very clear!
Hope questions with this ambiguity don't occur in the actual.
Kudos again!
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Re: A buyer buys 3 different items out of the newly introduced 10 differen [#permalink]

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Re: A buyer buys 3 different items out of the newly introduced 10 differen [#permalink]

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New post 16 Mar 2017, 09:58
Hello from the GMAT Club BumpBot!

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Re: A buyer buys 3 different items out of the newly introduced 10 differen [#permalink]

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New post 17 Mar 2017, 21:37
VeritasPrepKarishma wrote:
raghupara wrote:
Hi,

Someone pls correct me here...
P(choosing neither) = 1 - P(choosing either or both)

therefore,
P(ch. neither) = 1 - [(2C1.8C2 + 2C1.8C2 + 2C2.8C1)/10C3]
= 1 - (128/120).

Pls help.

Thanks


This is what the question means: (there is a debate on the intent due to the OA but this is what the question means to me)
Out of 10 items, a man buys 3. Now you have to pick any two items out of the original 10. What is the probability that both your items were not among those selected by the man too? (It is possible that one of your items match but both should not.)

P(both your items were not selected by man) = 1 - P(both items were selected by man) = 1 - 3C2/10C2 = 1 - 3/45 = 14/15

Intent as depicted by the OA:
What is the probability that neither of your items was among those selected by the man?

P(choosing neither of the man's items) = 7C2/10C2 = 21/45 = 7/15



hii mam , thanks for the previous reply , would you please correct me ...

i did this question like this :
probability of choosing item 1 . which is not from those 3 diff. items = 7/10
probability of choosing item 2,which is not from those 3 diff. items =6/9
so the answer will be 7/10*6/9=7/15

is my approach right ?
please guide
thanks
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A buyer buys 3 different items out of the newly introduced 10 differen [#permalink]

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New post 19 May 2017, 11:32
Bunuel wrote:
nphilli1 wrote:
A buyer buys 3 different items out of the newly introduced 10 different items. If two items were to be selected at random, what is the probability that the buyer does not have both the chosen items?

If possible, could people walk through steps and logic taken because I got an answer completely different than what was given.

Thanks

[Reveal] Spoiler:
Answer given was 7/15 with the logic of 7C2/10C2. IMO, that logic answers the question of the buyer not having either of the items. Am I wrong?


The answer 7/15 implies that the question means that buyer has none of the items selected - \(\frac{C^2_7}{C^2_{10}}=\frac{7}{15}\).

But "does not have both the chosen items" in my opinion should mean that buyer has none of the items selected OR has only one of the items selected BUT not both of them.

In this case: probability = 1 - probability of having both = \(1-\frac{C^2_3}{C^2_{10}}=\frac{14}{15}\).

Anyway not a good question.


Well, the question should certainly be reworded for clarity- OP do you have the official source for this question? More logically and fundamentally, the question should read "What is the probability that both of the items chosen DO NOT belong to the buyer? Anyways

7c2/10c2
A buyer buys 3 different items out of the newly introduced 10 differen   [#permalink] 19 May 2017, 11:32
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