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cards [#permalink] New post 05 Mar 2005, 21:38
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What is the probability of getting a King and a Queen in a single draw of two cards from a pack of well-shuffled cards?

1/16
(4c1*4c1)/(52c2)
(4*4)/(52*51)
(8c2)/(52c2)
56/(52c2)
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 [#permalink] New post 06 Mar 2005, 01:59
indipendent events so
prob of getting a king 4/52
prob of getting a queen 4/51 (no replacement)
4/52*4/51
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 [#permalink] New post 06 Mar 2005, 04:10
I'm a bit confused.
If order matters we have 4*4/(52*51)
But it seems that order doesnt matter!?
In that case it should be 4c1*4c1/52c2 (that is twice as much as the former result because Queen,King=King,Queen)
:?: :?: :?: :?: :?:
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Re: cards [#permalink] New post 06 Mar 2005, 05:55
MA wrote:
(4*4)/(52*51)


same way...same result :wink:

thearch, I don't see where the order is important or not...
For King and Queen we have the same number of possibilities, if it's king first and queen it will be the same than queen first and king second.

I admit that I didn't get your problem, please give more details... :roll:
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 [#permalink] New post 06 Mar 2005, 06:12
thearch wrote:
I'm a bit confused.
If order matters we have 4*4/(52*51)
But it seems that order doesnt matter!?
In that case it should be 4c1*4c1/52c2 (that is twice as much as the former result because Queen,King=King,Queen)
:?: :?: :?: :?: :?:


IMO it is 4c1*4c1/52c2 (order does not matter)

Solution when order matters in two or one way:

4/52*4/51 + 4/52*51 (order matters => king/queen or queen/king) or

4/52*4/51 (order matters => king/queen in that order)

Last edited by christoph on 06 Mar 2005, 12:37, edited 2 times in total.
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 [#permalink] New post 06 Mar 2005, 09:15
OA is

(4c1*4c1)/(52c2)
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Re: cards [#permalink] New post 06 Mar 2005, 09:49
Antmavel wrote:
thearch, I don't see where the order is important or not...
For King and Queen we have the same number of possibilities, if it's king first and queen it will be the same than queen first and king second.

I admit that I didn't get your problem, please give more details... :roll:

ok I agree but then the probability is 2*4*4/52*51 because K,Q=Q,K ->>the probability doubles
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 [#permalink] New post 06 Mar 2005, 14:22
cloaked_vessel wrote:
OA is

(4c1*4c1)/(52c2)


hmm... that's not same as 4*4/52*51. I guess that's because it's a single draw of two cards, not one card then the other. Can somebody elaborate pls?
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 [#permalink] New post 06 Mar 2005, 14:31
4c1*4c1/52c2

the question mentions that the two cards are drawn in a single draw and order does not matter....
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 [#permalink] New post 06 Mar 2005, 15:33
cloaked_vessel wrote:
OA is (4c1*4c1)/(52c2)

if you pick two cards at the same time not one by one.
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 [#permalink] New post 06 Mar 2005, 20:40
Even if you draw it one by one, if you don't replace the one you already picked before the second draw, would the answer be 4*4/52*51? Consider this: Are you sure you still have four choices when you do the second pick? What does the first 4 and the second 4 represent?
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 [#permalink] New post 07 Mar 2005, 02:09
Exactly, Hong. The first pick is either king or queen thus the second pick is not going to be effected by this.

c(4,1)*c(4,1)/c(52,2)
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 [#permalink] New post 07 Mar 2005, 07:30
In other words, the answer would be the same whether it is a single draw or two consequential draws without replacement.

The mistake for the answer 4/52*4/51 is this: The first draw you actually have 8 chances of drawing a correct card. Once you have drawn the K or Q, then you only have 4 chances to draw the remining correct card. Therefore the answer should be: 8/52*4/51, which turns out to be the same as C(4,1)*C(4,1)/C(52,2).

Of course you could also think as thearch did: The correct draws include a K and then a Q, or a Q and then a K. So the answer is 2*4/52*4/51.
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 [#permalink] New post 07 Mar 2005, 07:57
HongHu wrote:
In other words, the answer would be the same whether it is a single draw or two consequential draws without replacement.

The mistake for the answer 4/52*4/51 is this: The first draw you actually have 8 chances of drawing a correct card. Once you have drawn the K or Q, then you only have 4 chances to draw the remining correct card. Therefore the answer should be: 8/52*4/51, which turns out to be the same as C(4,1)*C(4,1)/C(52,2).

Of course you could also think as thearch did: The correct draws include a K and then a Q, or a Q and then a K. So the answer is 2*4/52*4/51.


i think this is not right when you say that we can also think of the question as Q and K or K and Q, because in this case the question should have a different wording: "What is the probability of getting a King and a Queen or a Queen and King in a single draw of two cards from a pack of well-shuffled cards?" :-D
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Re: cards [#permalink] New post 07 Mar 2005, 08:04
The question says "getting a King and a Queen in a single draw of two cards". That means the order doesn't matter. It is different from "getting a King and then a Queen when you draw two cards without replacement". In the latter the order matters. Basically what I'm trying to say is that the total outcome for something where the order doesn't matter is equivalent to the total outcomes for all the possible orders for something where the order does matter. :)
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 [#permalink] New post 07 Mar 2005, 08:19
RIGHT!

all possible orders: 4/52*4/51+4/52*4/51=16/1326

order doesnt matter: (4c1*4c1)/(52c2)=16/1326 or 8/52*4/51=16/1326 => but this one only works when the number of itmes is equal
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 [#permalink] New post 07 Mar 2005, 08:45
christoph wrote:
RIGHT!

all possible orders: 4/52*4/51+4/52*4/51=16/1326

order doesnt matter: (4c1*4c1)/(52c2)=16/1326 or 8/52*4/51=16/1326 => but this one only works when the number of itmes is equal


What do you mean by number of items are equal?
Using the combination formula should work, remember in combinations unlike permutations the reason why the formaula is n!/[(n-k)!*k!] is because the k! represents number of possibilities.
try it out, if you are chosing 3 cards, there are 6 ways (3!)
thats why they say in permutations order doesnt matter and hence the k! is not included. its just n!/(n-k)!
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 [#permalink] New post 07 Mar 2005, 09:07
Rupstar wrote:
christoph wrote:
RIGHT!

all possible orders: 4/52*4/51+4/52*4/51=16/1326

order doesnt matter: (4c1*4c1)/(52c2)=16/1326 or 8/52*4/51=16/1326 => but this one only works when the number of itmes is equal


What do you mean by number of items are equal?
Using the combination formula should work, remember in combinations unlike permutations the reason why the formaula is n!/[(n-k)!*k!] is because the k! represents number of possibilities.
try it out, if you are chosing 3 cards, there are 6 ways (3!)
thats why they say in permutations order doesnt matter and hence the k! is not included. its just n!/(n-k)!


assume there are 4 kings and 3 queens instead of 4 kings and 4 queens:

all possible orders: 4/52*3/51+3/52*4/51=12/1326

order doesnt matter: 4c1*3c1/52c2=12/1326 or 7/52*3/51 or 7/52*4/51=21/1326 => this way of calculations doesnt work for unequal items, so i would prefer the first approach

i just mentioned it because it could be like a shortcut for these questions, but i think its not inmportant.
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 [#permalink] New post 07 Mar 2005, 09:15
OK i get the picture,
its still the safest bet to trace the problem with the combination formula otherwise there is a chance of making a mistake and fogetting the combinations in which the cards can be drawn.
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 [#permalink] New post 15 Mar 2005, 10:11
great question and thanks everyone for the explanations.

i had one of those hunches that turned out to be correct but i could not really put my finger on it.

8/52*4/51 nice.
  [#permalink] 15 Mar 2005, 10:11
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