It is currently 22 Jun 2017, 17:33

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# cards

Author Message
Manager
Joined: 14 Dec 2004
Posts: 117

### Show Tags

05 Mar 2005, 22:38
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions

### HideShow timer Statistics

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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)
Senior Manager
Joined: 19 Feb 2005
Posts: 486
Location: Milan Italy

### Show Tags

06 Mar 2005, 02:59
indipendent events so
prob of getting a king 4/52
prob of getting a queen 4/51 (no replacement)
4/52*4/51
Senior Manager
Joined: 19 Feb 2005
Posts: 486
Location: Milan Italy

### Show Tags

06 Mar 2005, 05:10
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)
VP
Joined: 13 Jun 2004
Posts: 1115
Location: London, UK
Schools: Tuck'08

### Show Tags

06 Mar 2005, 06:55
MA wrote:
(4*4)/(52*51)

same way...same result

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.

VP
Joined: 30 Sep 2004
Posts: 1480
Location: Germany

### Show Tags

06 Mar 2005, 07:12
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, 13:37, edited 2 times in total.
Manager
Joined: 14 Dec 2004
Posts: 117

### Show Tags

06 Mar 2005, 10:15
OA is

(4c1*4c1)/(52c2)
Senior Manager
Joined: 19 Feb 2005
Posts: 486
Location: Milan Italy

### Show Tags

06 Mar 2005, 10:49
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.

ok I agree but then the probability is 2*4*4/52*51 because K,Q=Q,K ->>the probability doubles
Senior Manager
Joined: 02 Feb 2004
Posts: 344

### Show Tags

06 Mar 2005, 15:22
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?
Manager
Joined: 25 Oct 2004
Posts: 247

### Show Tags

06 Mar 2005, 15:31
4c1*4c1/52c2

the question mentions that the two cards are drawn in a single draw and order does not matter....
VP
Joined: 25 Nov 2004
Posts: 1483

### Show Tags

06 Mar 2005, 16:33
cloaked_vessel wrote:
OA is (4c1*4c1)/(52c2)

if you pick two cards at the same time not one by one.
SVP
Joined: 03 Jan 2005
Posts: 2233

### Show Tags

06 Mar 2005, 21:40
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?
Manager
Joined: 01 Jan 2005
Posts: 166
Location: NJ

### Show Tags

07 Mar 2005, 03:09
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)
SVP
Joined: 03 Jan 2005
Posts: 2233

### Show Tags

07 Mar 2005, 08:30
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.
VP
Joined: 30 Sep 2004
Posts: 1480
Location: Germany

### Show Tags

07 Mar 2005, 08:57
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?"
SVP
Joined: 03 Jan 2005
Posts: 2233

### Show Tags

07 Mar 2005, 09:04
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.
VP
Joined: 30 Sep 2004
Posts: 1480
Location: Germany

### Show Tags

07 Mar 2005, 09:19
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
Manager
Joined: 15 Feb 2005
Posts: 246
Location: Rockville

### Show Tags

07 Mar 2005, 09:45
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)!
VP
Joined: 30 Sep 2004
Posts: 1480
Location: Germany

### Show Tags

07 Mar 2005, 10:07
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.
Manager
Joined: 15 Feb 2005
Posts: 246
Location: Rockville

### Show Tags

07 Mar 2005, 10:15
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.
Manager
Joined: 11 Jan 2005
Posts: 57
Location: Mexico City

### Show Tags

15 Mar 2005, 11:11
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.
15 Mar 2005, 11:11
Display posts from previous: Sort by