<p>This wasn't an actual SAT Math II question, just a problem I came up with that's Math II material: There are 4 queens and 4 kings in a pack of 52 cards. If 5 cards are drawn at random without replacement, what is the probability that at least 1 queen AND at least 1 king are drawn? How would you do that?</p>
<p>Hmm. I’m not 100% about this, but:</p>
<p>Calculate the probability of drawing no kings and no queens, and then subtract that from one. </p>
<p>48/52 x 47/51 x 46/50 x 45/49 x 44/48 = probability of drawing no kings </p>
<p>48/52 x 47/51 x 46/50 x 45/49 x 44/48 = probability of drawing no queens</p>
<p>(48/52 x 47/51 x 46/50 x 45/49 x 44/48)^2 = probability of drawing no kings and no queens</p>
<p>1- (48/52 x 47/51 x 46/50 x 45/49 x 44/48)^2 = probability of drawing at least one king and one queen.</p>
<p>Someone want to corroborate this? Like I said, I’m not sure. </p>
<p>@Bemusedfyz careful. The complement of drawing at least one king and queen is drawing neither, OR drawing queens but no kings, OR drawing kings but no queen. Also the events of drawing no kings and drawing no queens are not independent.</p>
<p>There are 52P5 sequences of five cards we can draw, without replacement. (using notation nPr = n(n-1)…(n-r+1)).</p>
<p>How many do not contain a king or a queen? There are 44 cards that are neither king nor queen, so 44P5 sequences do not contain a king or a queen.</p>
<p>How many contain at least one king but no queens? There are 48 cards that are not queens, 48P5 sequences. Careful! This includes all five-card sequences with no kings. We have to subtract 44P5 to obtain the number of 5-card sequences with at least one king, but no queen. 48P5 - 44P5.</p>
<p>How many contain at least one queen but no king? This one is simple - we already figured this out (48P5 - 44P5 by symmetry).</p>
<p>Adding everything up, there are 2*48P5 - 44P5 5-card sequences that do not contain both queen and king, so dividing this by the total number of sequences and subtracting by 1, we obtain the probability:</p>
<p>P(>= 1 king, queen) = 1 - (2*48P5 - 44P5)/(52P5), about 10%.</p>
<p>–I’m about 99% sure this solution is correct but let me know if I made a mistake.</p>
<p>Yes, I thought that " The complement of drawing at least one king and queen is drawing neither, OR drawing queens but no kings, OR drawing kings but no queen" would be a problem, thank you for correcting me!</p>
<p>MITer94,</p>
<p>I think you’re correct, thanks (your answer agrees with another detailed answer on Yahoo Answers Math 
). But I’m going to really have to study your answer because probability is hard for me. Do you think this problem was too hard for Math II?</p>
<p>I think playing card problems are a really good way to hone probability skills.</p>
<p>Could you please explain why this is a permutation and not combination?</p>
<p>@cloudeleven it seems pretty tricky even for Math II, but it’s not a bad question by any means. Maybe it could be one of the harder Math II problems.</p>
<p>You may want to look at introductory counting/combinatorics books as well since those should go more in depth with permutations, combinations, as well as other topics such as inclusion-exclusion, stars/bars, etc…most of them should be very intuitive once you understand them.</p>
<p>I used nPr instead of nCr since the cards are drawn without replacement, so I decided to take order into account and treat them as sequences of five cards. Note that if I used nCr instead, the numerator/denominator would both be divided by 5! = 120 and we should obtain the same answer.</p>
<p>(note: I’ll sometimes avoid using the word “permutation” since a permutation is also a rearrangement of the elements of a set, e.g. 123, 132, 213, 231, 312, 321 are permutations of the set {1,2,3}).</p>
<p>Thanks.</p>