If you are drawing 3 cards onebyone with replacement from a
Solution
Total count of sample space S = 52C1 * 52C1 * 52C1 = (52)^3 = 140608
A (all 3 are red) = 26C1 * 26C1 * 26C1 = 17576 ways ( since there are two clubs of red cards)
B(all 3 are diamonds) = 13C1 * 13C1 * 13C1 = 2197 ways
C(all 3 are king of spades) = 1C1 * 1C1 * 1C1 = 1 way
D(all 3 are of the same color) = 26C1 * 26C1 * 26C1 * 2(either black or red colour) = 35152 ways
E(all 3 are from same suit) = 13C1 * 13C1 * 13C1 * 4 (there are four suits in total)
=> 8788 ways
F(all 3 of them are teh same cards) = 1 * 1 * 1 = 1 way
P(A) = 17576/140608 = 0.125
P(B) = 2197/140608 = 0.015625
P(C) = 1/140608 = 7.1197 * 10^(-6)
P(D) = 35152/140608 = 0.25
P(E) = 8788/140608 = 0.0625
P(F) = 1/140608 = 7.1197 * 10^(-6)
There are 52 - 12 = 40 number cards
G = 40C1 * 40C1 * 40C1 = 64000
P(G) = 64000/140608 = 0.4551
