# Three cards are selected from a deck of cards without replacement. What is the probability that you will get a king, a jack, and a queen?

Three cards are selected from a deck of cards without replacement. What is the probability that you will get a king, a jack, and a queen?
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Note that we can draw the first King(K), then Queen(Q) and then Jack(J). That means the sequence of drawings is (K,Q,J)(K,Q,J). Therefore we have 6 such a sequence of drawing cards: ​
(K,Q,J), (K,J,Q), (Q,K,J), (J,K,Q), (Q,J,K), (J,Q,K) ​
Note that each such events are equally likely. Thus in order to find the required answer we first consider one such case, say K,Q,J. Then the required probablity will be: $P=6\cdot P\left(K,Q,J\right)$
Let us first find the probability of drawing card first a King(K), then a Queen(Q) and then a Jack(J). ​
The probability of drawing a King in the 1st draw is Total Number of King / Total Number of Aviliable Cards $=4/52$
Since we can not replace the card, thus for the second draw we have total 51 cards available. ​
The probability of drawing a Queen in the 2nd draw is: Total Number of Queen / Total Number of Aviliable Cards $=4/51$
Since we can not replace the card, thus for the second draw we have total 50 cards available. ​
The probability of drawing a Jack in the 3rd draw is: Total Number of Jack / Total Number of Aviliable Cards $=4/50$
Thus the probability of drawing card first a King(K), then a Queen(Q) and then a Jack(J) is given by: ​
$P\left(K,Q,J\right)=\left(4/52\right)\cdot \left(4/51\right)\cdot \left(4/50\right)$
Hence the required probablity is: $6\cdot P\left(K,Q,J\right)=\frac{6\cdot {4}^{3}}{52\cdot 51\cdot 50}$