Question 31564: Find the number of different 5-card hands that can be dealt from a standard deck of 52 playing cards?
Find the probability that 4 kings and a queen are randomly drawn from a standard deck of 52 playing cards?
Found 2 solutions by longjonsilver, venugopalramana:
Answer by longjonsilver(2297)   (Show Source):
You can put this solution on YOUR website! ok....52 cards:
there are 52 possible choices...Pick one
there are now 51... Pick second
there are now 50... pick the third
there are now 49... pick 4th
there are now 48... pick 5th
So, possible number of variations are: 52*51*50*49*48. However, some of these are repeated. How many are repeated?
Well, look at an easier example: Pick 2 letters from ABCDE:
AB BA CA DA EA
AC BC CB DB EB
AD BD CD DC EC
AE BE CE DE ED
So from 5 choices we have 5*4 --> 20. And yet AB is the same as BA, in terms of them being playing cards...the hand is the same irrespective of having A then B dealt or B then A...the hand is still "A and B". Looking, there are actually only 10 different "hands" in this example: 20/2!.
Another example: pick 3 from the 5:
ABC BCD CDE
ABD BCE
ABE BDE
ACD
ACE
ADE
Here there are 10 solutions gotten from (5*4*3)/3!
So, for your question, we get 52*51*50*49*48/5!
which is 2,598,960 different hands
This gives us that the probability of a particular hand = 1/2598960
Now P(4 kings and a queen) is actually 4 potential hands...
4K and Q clubs or
4K and Q diamonds or
4K and Q hearts or
4K and Q spades
so, P(4K and a Q) = 4*(1/2598960)
4/2598960
As you can see, the odds are very small of this happening
jon.
Answer by venugopalramana(3286)  (Show Source):
You can put this solution on YOUR website! Find the number of different 5-card hands that can be dealt from a standard deck of 52 playing cards?
=52C5=52!/(52-5)!5!=52*51*50*49*48/5*4*3*2*1=52*51*10*49*2
Find the probability that 4 kings and a queen are randomly drawn from a standard deck of 52 playing cards?
4 KINGS CAN BE GOT FROM 4 KINGS IN =4C4=1 WAY
1 QUEEN CAN BE GOT FROM 4 QUEENS IN 4C1=4 WAYS.
SO NO.OF SUCCESSFULL COMBINATIONS =1*4=4
PROBABILITY OF SUCCESS =4/(52*51*10*49*2)=1/(52*51*5*49)
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