Find the opposite probability and subtract it from 1.

First card be any of the 12 cards, so P(1st card) = 12/12=1

Second card be any of the 10 cards out of the remaining 11 cards, so P(2nd card) = 10/11

Third card be any of the 8 cards out of the remaining 10 cards, so P(3rd card) = 8/10

Fourth card be any of the 6 cards out of the remaining 9 cards, so P(4th card) = 6/9

Probability of having no pair in the 4 card = 1*(10/11)*(8/10)*(6/9) = 16/33

Probability of having at least 1 pair = 1-(16/33) = 17/33

Ans: "C"

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Let me elaborate on one of the cases;

1,1,2,2,3,3,4,4,5,5,6,6 ->12 cards

1st card-> any card can be taken: 12/12=1

Let's say we picked 4 as the first card

2nd card be anything but 4

2nd card can be: 1,1,2,2,3,3,5,5,6,6 (Total Count=10)

Total available cards = 1,1,2,2,3,3,4,5,5,6,6(Total Count=11)

P(2nd Card) = 10/11

Let's say we picked 1 as the 2nd card.

3rd card be anything but 1 or 4

3rd card can be: 2,2,3,3,5,5,6,6 (Total Count=8)

Total available cards = 1,2,2,3,3,4,5,5,6,6(Total Count=10)

P(3rd Card) = 8/10

Let's say we picked 6 as the 3rd card.

4th card be anything but 1 or 4 or 6

4th card can be: 2,2,3,3,5,5(Total Count=6)

Total available cards = 1,2,2,3,3,4,5,5,6(Total Count=9)

P(4th Card) = 6/9

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~fluke

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