Probability - pairs of cards

sachin2411 wrote:A question from MGMAT CAT:

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, for each value from 1 to 6, there are two cards in the deck with that 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?

(A) 8/33

(B) 62/165

(C) 17/33

(D) 103/165

(E) 25/33

OA: (C) 17/33

This is how i calculated:
No. of ways of selecting a pair = 6C1

Once i have a pair, i am left with 10 cards,
So number of ways of selecting 2 from 10 = 10C2

So Probability = (6C1 X 10C2) / 14C4 [You considered only one pair, while the question is about at least one pair][ Note also that the 10C2 may or may not hold pairs, rendering your solution incorrect]

Please let me know, what is wrong with my approach.

YAMLAKSIRA wrote:

sachin2411 wrote:A question from MGMAT CAT:

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, for each value from 1 to 6, there are two cards in the deck with that 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?

(A) 8/33

(B) 62/165

(C) 17/33

(D) 103/165

(E) 25/33

OA: (C) 17/33

This is how i calculated:
No. of ways of selecting a pair = 6C1

Once i have a pair, i am left with 10 cards,
So number of ways of selecting 2 from 10 = 10C2

So Probability = (6C1 X 10C2) / 14C4 [You considered only one pair, while the question is about at least one pair][ Note also that the 10C2 may or may not hold pairs, rendering your solution incorrect]

Please let me know, what is wrong with my approach.

Well, 10C2 can be anything, either its a pair of 2 equal card or two different cards. At least 1 pair thing is already dealt by 6C1, for rest 2 it can be anything so 10C2.

Please help.

sachin2411 wrote:A question from MGMAT CAT:

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, for each value from 1 to 6, there are two cards in the deck with that 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?

(A) 8/33

(B) 62/165

(C) 17/33

(D) 103/165

(E) 25/33

OA: (C) 17/33

This is how i calculated:
No. of ways of selecting a pair = 6C1

Once i have a pair, i am left with 10 cards,
So number of ways of selecting 2 from 10 = 10C2

So Probability = (6C1 X 10C2) / 14C4

Please let me know, what is wrong with my approach.