The player gets 8 cards of 32. What is the probability that it gets
a) all 4 aces
b) at least 1 ace

Correct answer:

a =  0.1947 %
b =  70.4505 %

Step-by-step explanation:

C4(4)=(44)=4!(44)!4!=11=1  C4(28)=(428)=4!(284)!28!=432128272625=20475  C8(32)=(832)=8!(328)!32!=876543213231302928272625=10518300  a=100 (44) (832)(428)=899175%=0.1947%
C8(28)=(828)=8!(288)!28!=876543212827262524232221=3108105 b=100 (1(832)(8324))=70.4505%

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