From a pack of 52 cards..4 cards are drawn randomly..whats the probability that...?

Question

a)all of them are spade

b)one is of each suit

c)2 hearts and 2 diamonds

Answer 1

4 cards can be drwan in 52C4 ways.

52C4 = 52!/(4! * 48!)

a) All four spade can be selected in 13C4 ways.

Probability = 13C4/52C4

b) One of each suit = 13C1*13C1*13C1*13C1/52C4

c) 13C2*13C2/52C4

Answer 2

Unfortunately, all of the answers given so far are incorrect. They didn't think things though.

a)

There are (initially) 13 spades in (initially) 52 cards. After each draw, there is one less spade and the deck is smaller by 1:

13/52 * 12/51 * 11/50 * 10/49 = 17,160/6,497,400 = .00264

b)

For the first draw, it doesn't matter; any card will do

For the second draw, there is one less card in the deck, and any card in the remaining 3 suits will do.

For the third, there is yet one less card in the deck, and any card from the remaining 2 suits will do.

For the forth, there is again one less card in the deck, and you must pick a card from the one remaining suit:

52/52 * 39/51 * 26/50 *13/49 = 685,464/6,497,400 = .1055

c)

This one is more complicated.

There are 6 different ways to get 2 hearts and 2 diamonds; each combination has the same probability:

HHDD = 13/52 * 12/51 * 13/50 * 12/49 = (13*13*12*12)/(52*51*50*49)

HDHD = 13/52 * 13/51 * 12/50 * 12/49 = ...

HDDH = ...

DHDH = ...

DHHD = ...

DDHH = ...

So the probability of getting any of the above combinations is:

6*(13*13*12*12)/(52*51*50*49) = 6*26,336/6,487,400 = .0225

Answer 3

sample space = 52C4

a)

for all to be spades possibility is 13C4 (selecting 4 spades out of 13 spades)

so prob = (13C4)/(52C4)

b)

one form each suit so out of 4 groups fo 13 cards one to be drawn from each        13C1 * 13C1 * 13C1 * 13C1

to get prob divide above product by 52C4

c)

2 hearts out of 13 can be selected in 13C2 ways

2 diamonds out of 13 can be selected in 13C2 ways

so prob = (13C2 * 13C2)/(52C4)

Answer 4

a) it's (13/52)(12/51)(11/50)(10/49)

b) (13/52)(13/51)(13/50)(13/49)

c) (13/52)(12/51)(13/50)(12/49)

these are given that the cards b4 have came out the way you wanted

Answer 5

a] 13/52 * 12/51 * 11/50 *10/49

b] 13/51 * 13/50 * 13/49

c] 13/52 * 12/51 * 13/50 * 12/29

now just multiply the fractions

Answer 6

first you need to set the problem (s) up.  Here is an example to do the rest.  a.) 4 cards- all spades.  There are 13 total spades, 1-Ace.  For the purpose of this problem you must assume that each pick satisfies the parameters of the question.So the chance of getting a spade on the first pick = 13:52.  The chance of drawing a spade in pick two (without replacing the first pick) = 12:51,  the third pick = 11:50, fourth =10:49.  Now multiple each chance with each-other.  (13/52)X(12/51)X(11/50)X(10/49)= probability of drawing four consecutive spades from a deck of 52 cards.  Use your scientific calculator to solve.  

Answer 7

Only if I knew Magic, I could answer you!