With 2 card decks, I take 1 card from each one. What is the chance to draw a diamond at least once?
How about with a single deck with 21 cards removed (and you don't know which ones), and then 2 cards are drawn. What is the chance at least one of those is a diamond?
Hello, khor!
The second one has an incredibly long and tedious solution.
Since it depends on how many diamonds were removed,How about with one deck with 21 cards removed (you don't know which ones),
and then 2 cards are drawn. What is the chance at least one is a diamond?
. . we must consider all possible cases.
Suppose no is removed. . This probability is: .
The deck contains: 13 s and 18 Others.
The probability that no is drawn is: .
The probability that no is removed and at least one is drawn is: .
Suppose 1 is removed. . This probability is: .
The deck contains 12 s and 19 Others.
The probability that no is drawn is: .
The probabiity that 1 is removed and at least one is drawn is: .
. . .
We must extend this list up to:
Suppose 11 s are removed. . This probability is: .
The deck contains 2 s and 29 Others.
The probability that no is drawn is: .
The probability that 11 s are removed and at least one is drawn is: .
The final probability is the sum of these twelve probabilities.
You go on ahead . . . I'll wait in the car.
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After you carry out this computation, I think you will find the result is the same obtained by the simpler method I previously posted.
Look at it this way: You shuffle the deck and deal out the top two cards. Or you shuffle the deck, cut 21 cards from the top to the bottom, and deal out the top two cards. What's the difference? In terms of the distribution of the two cards dealt, none, really.