Bunuel wrote:
maheshsrini wrote:
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?
8/33
62/165
17/33
103/165
25/33
Let's calculate the opposite probability ans subtract this value from 1.
Opposite probability would be that there will be no pair in 4 cards, meaning that all 4 cards will be different: \(\frac{C^4_6*2^4}{C^4_{12}}=\frac{16}{33}\).
\(C^4_6\) - # of ways to choose 4 different cards out of 6 different values;
\(2^4\) - as each of 4 cards chosen can be of 2 different suits;
\(C^4_{12}\) - total # of ways to choose 4 cards out of 12.
So \(P=1-\frac{16}{33}=\frac{17}{33}\).
Or another way:We can choose any card for the first one - \(\frac{12}{12}\);
Next card can be any card but 1 of the value we'v already chosen - \(\frac{10}{11}\) (if we've picked 3, then there are one more 3 left and we can choose any but this one card out of 11 cards left);
Next card can be any card but 2 of the values we'v already chosen - \(\frac{8}{10}\) (if we've picked 3 and 5, then there are one 3 and one 5 left and we can choose any but these 2 cards out of 10 cards left);
Last card can be any card but 3 of the value we'v already chosen - \(\frac{6}{9}\);
\(P=\frac{12}{12}*\frac{10}{11}*\frac{8}{10}*\frac{6}{9}=\frac{16}{33}\).
So \(P=1-\frac{16}{33}=\frac{17}{33}\) - the same answer as above.
Answer: C.
Hope it helps.
Although I do understand your approach but not sure what's wrong with mine.!
I did this as follows -> P = P(one pair) + P(2 pairs)
6C1*10C1*8C1/12C4 + 6C2/12C4
6C1 - no. of ways of selecting 1 pair out of 6 pairs
10C1 - no. of ways of selecting 1 card out of the remaining 10
8C1 - no. of ways of selecting 1 card out of the remaining 8 cards ( excluding the 9th card as that would form another pair which is considered in the second case)
12C4 - total no. of ways of selecting 4 cards out of 12 cards
6C2 - no. of ways of selecting 2 pairs out of 6 pairs
I am not getting the right answer - not sure which point I am missing. Can you please explain?
Thanks.!