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MGMAT Q25
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Cybermusings
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jayhawk2001
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Hmm, this one had me going for a while.
Total number of possibilities = 12 * 11 * 10 * 9
Out of this, no-pair = 12 * 10 * 8 * 6
So, prob of no-pair = 12*10*8*6 / 12*11*10*9 = 16/33
So, prob of atleast 1 pair = 1 - 16/33 = 17/33
...Alternatively...
Now, I can't seem to solve this the straight way.
Num possibilities for 2 pair = 12 * 6 * 10 * 5
Num possibilities for 1 pair (only) = 12 * 6 * 10 * 8
Summing the above, I get 12*6*10*13 / 12*11*10*9 = 26/33
I'm definitely missing something.
Total number of possibilities = 12 * 11 * 10 * 9
Out of this, no-pair = 12 * 10 * 8 * 6
So, prob of no-pair = 12*10*8*6 / 12*11*10*9 = 16/33
So, prob of atleast 1 pair = 1 - 16/33 = 17/33
...Alternatively...
Now, I can't seem to solve this the straight way.
Num possibilities for 2 pair = 12 * 6 * 10 * 5
Num possibilities for 1 pair (only) = 12 * 6 * 10 * 8
Summing the above, I get 12*6*10*13 / 12*11*10*9 = 26/33
I'm definitely missing something.
