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Fermat’s card problem. Consider a 40-card deck with four each of ten denominations (e.g., one could...

Fermat’s card problem. Consider a 40-card deck with four each of ten
denominations (e.g., one could eliminate the 8s, 9s, and 10s from the 52-card
deck). What is the probability that four cards, drawn at random without
replacement, are of different suits?

Math   Statistics-And-Probability   
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Answer #1

Total number of card = 40

Cards of each suit = 10

P[ drawn at random without replacement, are of different suits ] = selecting 1 card from each suit / selecting 4 cards out of 40

selecting 1 card from each suit = 4*10C1 ( 4 suits )

selecting 4 cards out of 40 = 40C4

P[ drawn at random without replacement, are of different suits ] = 4*10C1 / 40C4

P[ drawn at random without replacement, are of different suits ] = 4*10 / 91390

P[ drawn at random without replacement, are of different suits ] = 40 / 91390

P[ drawn at random without replacement, are of different suits ] = 0.00043768

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