nonameee wrote:

Four cards are randomly drawn from a pack of 52 cards. Find the probability that all four cards are of different denominations?

My solution:

OMEGA = 52 x 51 x 50 x 49

A = 52 x 48 x 44 x 40

P(A) = A/OMEGA = 0.68

Official solution:

OMEGA: (52 4)

A: (4 1) x (13 4)

P (A) = 0.01

I think the official solution is wrong.

nonameee wrote:

nonameee wrote:

My solution:

OMEGA = 52 x 51 x 50 x 49

A = 52 x 48 x 44 x 40

P(A) = A/OMEGA = 0.68

I sorted it out. My solution is wrong: 52 x 48 x 44 x 40 means that this set will actually contain cards of the same denomination. The official solution is IMO still incorrect since A should be equal to: (4^4) x (13 4).

Your opinion would be greatly appreciated.

I think you are right:

\(\frac{4^4*C^4_{13}}{C^4_{52}}\) would be the probability of getting 4 cards of different denominations out of 52.

# of ways of to choose 4 different denominations out of 13 -\(C^4_{13}\);

Multiplying by \(4^4\) as each card from the above string can be of four suits (\(4*4*4*4=4^4\));

Total # of ways to choose 4 cards out of 52 - \(C^4_{52}\).

Or another way:We can choose any card for the first one - \(\frac{52}{52}\);

Next card can be any card but 3 of the denomination we'v already chosen - \(\frac{48}{51}\) (if we've picked 10, then there are 3 10-s left and we can choose any but these 3 cards out of 51 cards left);

Next card can be any card but 3*2=6 of the denominations we'v already chosen - \(\frac{44}{50}\) (if we've picked 10 and King, then there are 3 10-s and 3 Kings left and we can choose any but these 3*2=6 cards out of 50 cards left);

Last card can be any card but 3*3=9 of the denominations we'v already chosen - \(\frac{40}{49}\);

\(P=\frac{52}{52}*\frac{48}{51}*\frac{44}{50}*\frac{40}{49}=\frac{4^4*13*12*11*10}{52*51*50*49}\) - the same answer as above.

Hope it helps.

No posting of PS/DS questions is allowed in the main Math forum.[/warning]

In your solution, you found the # of ways to choose 4 different denominations out of 13, I believe that is incorrect. Each suit(denomination) has 13 cards, we're looking for the # of ways to draw 1 card of a different suit each time, up to 4 draws. Correct me if I'm wrong.