Homework Answers
3)a) P(All cards will be aces) = 4C4/52C4
= 0.000004
b) P(There will be no aces) = 48C4/52C4
= 0.718737
C) P(All 4 will be one suit) = 4 * 13C4/52C4
= 0.010564
d) P(All cards will be same colour) = 2 * 26C4/52C4
= 0.110444
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