Asked by B.B.
Three cards are drawn without replacement from an ordinary deck of 52 playing cards. What is the probability that the second and third cards are spades if the first card was not a spade?
Answer: 36/52 18/26 9/13= 69%.
Is this right? Thanks for the help.
Answer: 36/52 18/26 9/13= 69%.
Is this right? Thanks for the help.
Answers
Answered by
B.B.
Wrong question please don't answer. Thanks.
Answered by
drwls
No, that is not right. Doesn't 69% seem too high a probability for drawing spades twice in a row?
This is a conditional probability. A key phrase is "...IF the first card was not a spade." You have to assume that the first card was NOT a spade. Then the deck before the second draw has 13 spades out of 52. The same applies to the third draw, since the drawn spades are replaced. The answer is (13/52)^2 = 1/16
This is a conditional probability. A key phrase is "...IF the first card was not a spade." You have to assume that the first card was NOT a spade. Then the deck before the second draw has 13 spades out of 52. The same applies to the third draw, since the drawn spades are replaced. The answer is (13/52)^2 = 1/16
Answered by
MathMate
If we consider "without replacement", the second draw starts with only 51 cards and 13 spades.
The probability of success would be the product of the probabilities of the second and third draws, namely
(13/51)*(12/50)=26/425
The probability of success would be the product of the probabilities of the second and third draws, namely
(13/51)*(12/50)=26/425
Answered by
drwls
I did the with replacement case by mistake. Careless of me. Sorry
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