Suppose you are dealt six cards from a standard deck of 52 playing
cards. What is the probability of being dealt exactly 3 spades and
3 fives? What is then the probability of being dealt at least 3
spades and 3 fives?
First we calculate the number of successful hands
Case 1: The 5S is not included.
Choose the ranks for the three spades in 12C3 = 220 ways.
Choose the suits for the three fives in 3C3 = 1 way
That's 220*1 = 220 ways for case 1
Case 2: The 5 of spades is included.
We choose the 5 of spades 1C1 = 1 way.
We choose the ranks for the other 2 spades in 12C2 = 66 ways.
We choose the suits for the other 2 fives in 3C2 = 3 ways
We choose the one card that is neither a spade nor a five
in 36C1 = 36 ways.
[It's 36 because 52 cards - 13 spades - 3 non-spade 5's = 36]
That's 1*66*6*36 = 14256 for case 2
Total successful hands = 220 + 14256 = 14476
Total possible hands = 52C6 = 20358520
Probability = 14476/20358520 which reduces to 11/15470.
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What is then the probability of being dealt at least 3 spades
and 3 fives?
There is only two additional case besides the preceding,
1. the case where all 4 5's are chosen, including the 5 of spades,
and 2 more spades.
Choose the 4 5's 4C4 = 1 way
Choose the ranks for the other 2 spades 12C2 = 66 ways
That's 1*66 = 66 ways
2. the case where 4 spades are chosen, including the 5 of spades,
and 2 more 5's.
Choose the 5 of spades 1 way,
Choose the ranks for the other 3 spades 12C3 = 220 ways
Choose the other 2 5's 3C2 = 3 ways.
That's 1*220*3 = 660 ways.
That's an additional 66+660 = 726 ways to add to the 14476
from the preceding problem, or 14476+726 = 15202 ways.
Total possible hands = 52C6 = 20358520
Probability = 15202/20358520 which reduces to 7601/10179260.
Edwin