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06 Aug 2023, 09:56
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solo1234
A drawer has six loose blue socks and six loose white socks. If four socks are removed from the drawer at random and without replacement. What is the probability that one pair of each color was selected?
A. 2/33
B. 5/66
C. 5/33
D. 5/11
E. 1/2
A. 2/33
B. 5/66
C. 5/33
D. 5/11
E. 1/2
C(2)6*C(2)6/C(4)12=5/11
Posted from my mobile device
lecremeglace
GMAT 1: 730 Q49 V42
Posts: 25
06 Aug 2023, 17:00
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VedantChourey
Hi Mod's or anyone, whats wrong with this approach
I SOLVED THIS QUESTION IN THIS WAY,
6/12*5/11*6/10*5/9
First getting any 6 blue ball out of 12, then getting 5 blue ball(because its mentioned no replacement) out of 11, then getting any 6 white ball out of 10 and at last getting any 5 white ball out of 9. It gave me 5/66.
Whats wrong with this approach.
I SOLVED THIS QUESTION IN THIS WAY,
6/12*5/11*6/10*5/9
First getting any 6 blue ball out of 12, then getting 5 blue ball(because its mentioned no replacement) out of 11, then getting any 6 white ball out of 10 and at last getting any 5 white ball out of 9. It gave me 5/66.
Whats wrong with this approach.
Hi VedantChourey, I think you're on the right track if you wanted to use the probability method on this question.
However, you've only considered one case, that is Blue-White-Blue-White. You must consider all cases if you want to use this method. Keep in mind that it can be time consuming (it took me over 2min to come up with the cases and write out the formulas).
Double your first equation as the option of WWBB exists, giving us \(\frac{10}{66}\).
BWWB results in \(\frac{6}{12}*\frac{6}{11}*\frac{5}{10}*\frac{5}{9}\), giving another \(\frac{5}{66}\) or \(\frac{10}{66}\) if we consider the reverse (WBBW).
Finally, we have BWBW, which is the same formula as above, doubled to \(\frac{10}{66}\) for the reverse (WBWB).
In total, that is \(\frac{10}{66}\) three times, which gives us \(\frac{10}{66}*3=\frac{5}{11}\). Option D.
Hope that helps!
12 Sep 2024, 00:25
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VedantChourey
Hi Mod's or anyone, whats wrong with this approach
I SOLVED THIS QUESTION IN THIS WAY,
6/12*5/11*6/10*5/9
First getting any 6 blue ball out of 12, then getting 5 blue ball(because its mentioned no replacement) out of 11, then getting any 6 white ball out of 10 and at last getting any 5 white ball out of 9. It gave me 5/66.
Whats wrong with this approach.
I did the same but adjusted for the "repetitions", so one possibility is BBWW, adjusting for the "repeated" colour socks gives us 4!/2!*2! permutations which is 6, multiply this by ONE possible order of the favorable probability, so we have 6*5/66 which yields 5/11 I SOLVED THIS QUESTION IN THIS WAY,
6/12*5/11*6/10*5/9
First getting any 6 blue ball out of 12, then getting 5 blue ball(because its mentioned no replacement) out of 11, then getting any 6 white ball out of 10 and at last getting any 5 white ball out of 9. It gave me 5/66.
Whats wrong with this approach.
