Three cards are selected at random without replacement from a well-shuffled deck of 52 playing cards. Find the probability of drawing three cards of the same suit. I got .235 but I'm supposed to round to 5 decimal places. Am I right? How would I do this problem?

Question

anonymous · Answer

I believe the way you would solve this would be: 
13/52 times 12/52 times 11/52

Because the first time you have a 13/52 shot it would be in that suit, then the second time there are 12 more cards of that suit left, then the third time there are only 11 cards of that suit left.

anonymous · Answer

I tried that and typed it in. It was wrong and it was my last try on that problem. Thanks though!

anonymous · Answer

@psychotink, it's without replacement, so that'll be 12/51 and 11/50. Besides, there are 4 possibilities of this happening, dependeing on the suit you picked

campbell_st · Answer

if there is no replacement 

1st draw = 13/52

2nd draw   12/51

3rd draw 11/50

the probability is the product of the 3 draws

anonymous · Answer

Tried that too. It didn't work. Thanks though!

anonymous · Answer

have you tried (13/52*12/51*10/50) * 4?

campbell_st · Answer

how many are drawn... only 3

anonymous · Answer

yeah, but you have P(all hearts)+P(all clubs)+P(All spades)+P(all diamonds). P of any of those are, as you said, (13/52)*(12/51)*(11/50)

anonymous · Answer

I think it was *4. All my tries are used up though so I won't know the right answer. I tried (13/52)*(12/51)*(11/50) and (13/52)+(12/51)+(11/50), neither of them worked.

anonymous · Answer

ah, that's too bad