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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.
 one year ago
 one year ago
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.
 one year ago
 one year ago

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Nick2019Best ResponseYou've already chosen the best response.0
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 !!)
 one year ago

HesanBest ResponseYou've already chosen the best response.0
i m getting the wrong answer
 one year ago

terenzreignzBest ResponseYou've already chosen the best response.0
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?
 one year ago

BenjamminBest ResponseYou've already chosen the best response.0
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%
 one year ago

HesanBest ResponseYou've already chosen the best response.0
@Benjammin: I did the same thing but getting the wrong answer
 one year ago

ZarkonBest ResponseYou've already chosen the best response.1
are you supposed to get one of each card..one king and one queen?
 one year ago

HesanBest ResponseYou've already chosen the best response.0
@Benjammin: your calculations give the answer: 14/663
 one year ago

ZarkonBest ResponseYou've already chosen the best response.1
\[\frac{{4\choose 1}{4\choose1}}{{52\choose 2}}\]
 one year ago

ZarkonBest ResponseYou've already chosen the best response.1
I gave the formula above
 one year ago

ZarkonBest ResponseYou've already chosen the best response.1
this is a classic hypergeometric distribution problem
 one year ago

ZarkonBest ResponseYou've already chosen the best response.1
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 \[2\cdot\frac{4}{52}\cdot\frac{4}{51}\]
 one year ago

HesanBest ResponseYou've already chosen the best response.0
but there shoulb be 8 cards right? 4 for kings and 4 for queens
 one year ago

ZarkonBest ResponseYou've already chosen the best response.1
yes..hence the two 4's in the solution
 one year ago

HesanBest ResponseYou've already chosen the best response.0
okey, means 8/52, now how did u get 4/51?
 one year ago

SmoothMathBest ResponseYou've already chosen the best response.1
The way I see it, it's simply: 8/52 * 4/51
 one year ago

ZarkonBest ResponseYou've already chosen the best response.1
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.
 one year ago

ZarkonBest ResponseYou've already chosen the best response.1
pick from the 8...then don't pick a matching card...hence only 4 left to choose from
 one year ago

SmoothMathBest ResponseYou've already chosen the best response.1
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.
 one year ago

HesanBest ResponseYou've already chosen the best response.0
no i dont get it.. i was thinking it 7/51
 one year ago

HesanBest ResponseYou've already chosen the best response.0
see we've drawn 1 card, now 7 are left, so it would be 7/51
 one year ago

ZarkonBest ResponseYou've already chosen the best response.1
but by doing that you could get a matching pair...2 kings or 2 queens
 one year ago

HesanBest ResponseYou've already chosen the best response.0
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?
 one year ago

ZarkonBest ResponseYou've already chosen the best response.1
that would work...that is what SmoothMath did
 one year ago

HesanBest ResponseYou've already chosen the best response.0
if we take 4 cards for 2nd turn, will it not be 49 in the denominator? 4/48
 one year ago

BenjamminBest ResponseYou've already chosen the best response.0
Oooh, I see. What I gave was to draw two cards, each of which would be a king or queen.
 one year ago

ZarkonBest ResponseYou've already chosen the best response.1
@Hesan no the experiment is drawing 2 cards from 52....so it is 52 then 51 as above
 one year ago

HesanBest ResponseYou've already chosen the best response.0
okey, now what is the probability that the cards are both jack?
 one year ago

HesanBest ResponseYou've already chosen the best response.0
the second will always be /51 if we draw 2 cards?
 one year ago

ZarkonBest ResponseYou've already chosen the best response.1
yes...if you are drawing without replacement
 one year ago

ZarkonBest ResponseYou've already chosen the best response.1
because you pick one of the 52 cards...there are only 51 cards remaining
 one year ago

HesanBest ResponseYou've already chosen the best response.0
alright, thanks.. i get it now!
 one year ago
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