- #1
Frank69
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Hello people. I m new to probability and now I have to give an exam at uni about it. One of the sample question is like this:
Four cards are drawn from a pack of 52 well-shuffled cards. Find the probability that 3 are aces and 1 is a king.The way I see it, it should be like:
A = { (ace)(ace)(ace)(king), (ace)(ace)(king)(ace), (ace)(king)(ace)(ace), (king)(ace)(ace)(ace) }
P(A) = P((ace)(ace)(ace)(king)) + P((ace)(ace)(king)(ace)) + P((ace)(king)(ace)(ace)) + P((king)(ace)(ace)(ace))
Where:
P((ace)(ace)(ace)(king)) = (4/52*3/51*2/50*4/49)
P((ace)(ace)(king)(ace)) = (4/52*3/51*4/50*2/49)
P((ace)(king)(ace)(ace)) = (4/52*4/51*3/50*2/49)
P((king)(ace)(ace)(ace)) = (4/52*4/51*3/50*2/49)
is this correct?
if not, would you show me the proper way to solve such problems?
Sorry for the very basic question, but I m real noob.
Thanks in advance
Four cards are drawn from a pack of 52 well-shuffled cards. Find the probability that 3 are aces and 1 is a king.The way I see it, it should be like:
A = { (ace)(ace)(ace)(king), (ace)(ace)(king)(ace), (ace)(king)(ace)(ace), (king)(ace)(ace)(ace) }
P(A) = P((ace)(ace)(ace)(king)) + P((ace)(ace)(king)(ace)) + P((ace)(king)(ace)(ace)) + P((king)(ace)(ace)(ace))
Where:
P((ace)(ace)(ace)(king)) = (4/52*3/51*2/50*4/49)
P((ace)(ace)(king)(ace)) = (4/52*3/51*4/50*2/49)
P((ace)(king)(ace)(ace)) = (4/52*4/51*3/50*2/49)
P((king)(ace)(ace)(ace)) = (4/52*4/51*3/50*2/49)
is this correct?
if not, would you show me the proper way to solve such problems?
Sorry for the very basic question, but I m real noob.
Thanks in advance
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