Suppose you are dealt a hand of 3 cards at random from a deck of 24
cards consisting of Four Aces, Four 2s, Four 3s, and so on up to Four 6s.
i) What is the probability that your hand will have at least 2 hearts?
ii) What is the probability that your hand will have at least 2 cards in
the same suit?
Total number of 3-card hands is:
24 × 23 × 22 / 3! = 4 × 23 × 22 = 2024
Call that number T.
i) What is the probability that your hand will have at least 2 hearts?
There are 6 of each suit.
There 18 non-hearts so the number of ways of getting 0 hearts is as follows:
18 × 17 × 16 / 3! = 18 × 17 × 16 / 6 = 3 × 17 × 16 = 816
So chances of 0 hearts is 3 × 17 × 16 / T.
If the first card is a heart and the other two not,
it's 6 × 18 × 17, and similarly for 2nd or 3rd card being a heart.
So 3 × (6 × 18 × 17) ways of getting 1 heart.
But then we must factor out that we don't care about the order after all:
18 × 18 × 17 / 3!
or 3 × 18 × 17 = 918
So the chances of getting 0 or 1 hearts is
P = (3 × 17 × 16) + (3 × 18 × 17) all divided by T. (816+918)/2024 = 85.67%
And the chances of getting at least 2 hearts is 1 - P = 290/2024 = 14.33%.
ii) What is the probability that your hand will have at least 2 cards in
the same suit?
Let Q = chances of all three from different suits
To get all three from different suits we have:
24 × 18 × 12 / 3! ways or 4 × 18 × 12 = 864
So Q = 4 × 18 × 12 / T = 42.69%
then the answer is 1 - Q = 57.31%
