Hi,

Three cards are drawn from a pack of 52, 

"each card being replaced before the next one is drawn"

P(all spades) = P(1st one a spade 'and' 2nd one a spade 'and' 3rd one a spade)

              = 13/52*13/52*13/52 = 1/4*1/4*1/4 = 1/64 = .0156 OR 1.56%

"Without replacememt!"

P(all spades) = 13/52*12/51*11/50 = 1716/132,600 = 286/22,100 = .0129 OR 1.29%



What that says 'statistically' is that replacing the card...ups the chance of

all spades being drawn 

Answer by stanbon(75887)   (Show Source): You can put this solution on YOUR website!
 Three cards are drawn from a pack of 52, each card being replaced before the next one is drawn. Compute the probability p that all are spades. A pack of 52 cards includes 13 spades. 

Here is how I did it: 

p = 13C3 / 52C3 = 286/22,100 = 0.0129 or 1.29%.

---

Comment: Your's is the probability of picking 

a set of 3 clubs from all the sets of 3 cards.

--------------------------------------

 However, this is NOT the right answer. My textbook's solution is solved like this: (13/52)^3 = 1/64 = 0.0156 or 1.56%.

Comment: This is the probability of successively 

selecting a club 3 times from a deck of 52 cards.

-------------------------------------

Comment: An interesting difference. 

==============

cheers,

Stan H. 

 

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