probability - cards

[manasdas]

Bill has a small deck of 12 playing cards made up of only 2 suits of 6 cards each. Each of the 6 cards within a suit has a different value from 1 to 6; thus, there are 2 cards in the deck that have the same value.
Bill likes to play a game in which he shuffles the deck, turns over 4 cards, and looks for pairs of cards that have the same value. What is the chance that Bill finds at least one pair of cards that have the same value?

Please show thy workings..

[e.cartman]

deleted

[e.cartman]

Here is attempt 2:
For not choosing any pair,
12*10*8*6/(12*11*10*9)
=16/33.

I am assuming first card can be picked in 12 ways, second different value in 10 ways, third different value in 8 ways and fourth in 6 ways.
So for choosing at least 1 pair,
prob = 1 - 16/33 = 17/33.

Please post Official Answer.

[amitabh007]

.....is it 1/6
total pairs..11,12,13,14,15,16...,21,22,23,24,25,26,.... 31,32,33,34,35,36,......a total of 36 pairs
favourable...11,22,33,44,55,66...a total of 6 pairs
prob of at least 1 favo pair = 6/36*30/35+6/36*5/35
this comes out to be 1/6
Official Answer PLS...

[supersuj]

Answer is 17/33

-Select one card, probability is 1

-Select card two, probability that it is not the same number as card one is 10/11

-Select card three, probability that it is not the same number as card one or card two is 8/10

-Select card four, probability that it is not the same number as card one, two or three is 6/9

Therefore, probability that all 4 cards have a different number is
1*(10/11)*(8/10)*(6/9)= 16/33

Probability that there is atleast one pair is 1-16/33 = 17/33

[megha57]

imo 17/33

[manasdas]

as explained, Official Answer is 17/33

[namitb]

another take at it is

we can choose either suit in 1/2 ways

now in order that there is no pair we need to select all the 4 from card from one pair
so probability = 1/2(6c4/12c4) = 16/33
1- the probability that no pair is selected = the probability thats atleast one pair is selected
thus 17/33

[bose]

got the same ans 17/33