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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
If possible, could people walk through steps and logic taken because I got an answer completely different than what was given.
Thanks
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?
27 May 2010, 11:54
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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
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
27 May 2010, 12:01
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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
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
