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Two cards are drawn at random from a deck of well shuffled cards. Determine the probability that both the cards are a king and a queen.

Mathematics
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no
two cards, remember
there are 4 suites of 13 cards each , so the probability of getting two cards in each suit is 2/13 there are four suits ,so the probability is 2\13*4 = 8\13 -(use calculator to make it to decimals !!)

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Other answers:

i m getting the wrong answer
I'm not getting the question :( The two cards, are they supposed to be both kings or both queens or one king and one queen?
Well, there are 52 cards in a deck and 8 of them are Kings and Queens. So the probability of one random draw is 8/52. Assuming you draw one, there are 7/51 remaining, so you just need to multiply that out to get the answer. |dw:1341587716616:dw| ... by my calculations, about 2.11%
@Benjammin: I did the same thing but getting the wrong answer
are you supposed to get one of each card..one king and one queen?
@Benjammin: your calculations give the answer: 14/663
@Zarkon: yes
\[\frac{{4\choose 1}{4\choose1}}{{52\choose 2}}\]
\[\frac{8}{663}\]
yeah right.. How?
I gave the formula above
this is a classic hypergeometric distribution problem
you can look at it this way \(\frac{4}{52}\) for the first card and \(\frac{4}{51}\) for the second card but the order we pick them doesn't matter so the answer is \[2\cdot\frac{4}{52}\cdot\frac{4}{51}\]
but there shoulb be 8 cards right? 4 for kings and 4 for queens
yes..hence the two 4's in the solution
okey, means 8/52, now how did u get 4/51?
The way I see it, it's simply: 8/52 * 4/51
if you choose the cards from the 8 (4K and 4Q) you might pick up 2 kings or 2 queens, but you only want one of each.
yes...same thing
pick from the 8...then don't pick a matching card...hence only 4 left to choose from
Right, for the first card, a king OR a queen is a good outcome. However, assuming you pull one of those, for the second card, only one of those would be a good outcome.
no i dont get it.. i was thinking it 7/51
see we've drawn 1 card, now 7 are left, so it would be 7/51
but by doing that you could get a matching pair...2 kings or 2 queens
fine, suppose that on drawing first card, we've got a queen, so we'll exclude queen and now we'll take 4 kings.. am i right?
that would work...that is what SmoothMath did
right. one thing more..
if we take 4 cards for 2nd turn, will it not be 49 in the denominator? 4/48
Oooh, I see. What I gave was to draw two cards, each of which would be a king or queen.
@Hesan no the experiment is drawing 2 cards from 52....so it is 52 then 51 as above
i m getting it now...
okey, now what is the probability that the cards are both jack?
(4*52)*(3*51)
the second will always be /51 if we draw 2 cards?
yes...if you are drawing without replacement
because you pick one of the 52 cards...there are only 51 cards remaining
alright, thanks.. i get it now!

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