Posted by B.B. on Monday, August 3, 2009 at 8:03am.
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.

Math  B.B., Monday, August 3, 2009 at 8:38am
Wrong question please don't answer. Thanks.

Math  drwls, Monday, August 3, 2009 at 8:39am
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

Math  MathMate, Monday, August 3, 2009 at 8:44am
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

Math  drwls, Monday, August 3, 2009 at 10:51am
I did the with replacement case by mistake. Careless of me. Sorry
Answer This Question
Related Questions
 Math  Two cards are drawn without replacement from an ordinary deck of 52 ...
 math 157  three cards are drawn without replacement from an ordinary deck of 52...
 Math  Two cards are drawn without replacement from an ordinary deck of 52 ...
 math  Two cards are drawn without replacement from an ordinary deck of 52 ...
 Math/ Probability  Two cards are drawn without replacement from an ordinary ...
 probability/math  Three cards are drawn without replacement from an ordinary ...
 math  2 cards are drawn without replacement from an ordinary deck of 52 playing...
 math  Two cards are drawn without replacement from an ordinary deck of 52 ...
 Math check answer  Two cards are drawn without replacement from an ordinary ...
 Math check answer  Two cards are drawn without replacement from an ordinary ...
More Related Questions