Bill has a small deck of 12 playing cards

[tim]

aside from the fact that your notation is backwards, it looks like everything checks out on your solution. definitely overkill as compared to just using numbers, but it certainly works..

[NMencia09]

use slot method... and (1-x) strategy.

probability of getting a pair= 1 - probability of not getting pair...

where not getting pair means they are all different = 12 X 10 X 8 X 6 = 5760.

total # ways to choose 4 cards from 12 = 12X11X10X9 = 11880.

1 - (5760/11880) = .515151 which is the same as 17/33.

[tim]

thanks. let us know if there are any further questions on this one..

[fridaymusa]

use slot method... and (1-x) strategy.

probability of getting a pair= 1 - probability of not getting pair...

where not getting pair means they are all different = 12 X 10 X 8 X 6 = 5760.

total # ways to choose 4 cards from 12 = 12X11X10X9 = 11880.

1 - (5760/11880) = .515151 which is the same as 17/33.

N:B: 12x10c8x6 / 12x11x10x9 = 8x6/11x9 = 16/33
So, 1 - 16/33 = 17/33(desired ans)

[RonPurewal]

fridaymusa, do you have a question?
your post appears to be the same method as the one you're quoting (though with a bit more simplification of the fraction).

if you have a question, please clarify. thanks.

[vikramd198]

Hi. Since it says find probability of at least one pair, shouldn't we add 6C2 + 6C1*10C2 (instead of subtracting), in which 6C2 is the case when 2 pairs are drawn, and 6C1*10C2 is the case when only 1 pair is drawn and the other 2 are different?? Would be very grateful if you could clear this concept for me.

[RonPurewal]

sorry, i can't really tell what your whole work-up is, here. how are you using these calculations? what do they represent? what will you divide them by (to produce a probability expression)?

also, you wrote "instead of subtracting" -- instead of subtracting what? are you referring to ... an answer key? some other post on this thread?

please clarify, thanks.