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MDM4U Probability Question?
I have no idea how to solve this question. Every answer I get is wrong. Thanks in advance.
In the game of rummy, a player wins with a hand of three-of-a-kind or arun of three or more cards in the same suit. What is the probability that a hand of seven cards will be dealta) three kings?b) three-of-a-kind?c) the 4, 5, and 6 of spades?d) a run of exactly three cards?
1 Answer
- nbsale (Freond)Lv 64 weeks ago
You first find the number of combinations that give you the specified portion of the hand, and then multiply by the number of ways to fill out the hand. I'll do one of the cases, 3 kings.
To start, you need to decide if that is exactly 3 kings, or if you could have 4 kings. You probably want to calculate each case separately. There is only 1 way to get 4 kings. Then there are 48 other cards, and there are 48x47x46 permutations for the other 3 cards in the hand, and 3! = 6 ways to permute them. So there are (1)x48x47x46/6 possible hands.
Now for the case of exactly 3 kings, there are 4C3 = 4 combinations of the kings. For the other 4 cards, there are 48x47x46x45 / 4! combinations.
Now take the sum of those, and divide by the total number of possible hands, 52C7.
That's the general idea. It will take some thought on how to break up the others into cases.
