Probability - Aces, 2s, 3s, 4s
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?
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1 solution
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%