SOLUTION: Two cards are drawn in succession without replacement from a standard deck of 52 playing cards. What is the probability that the first card drawn was a diamond given that the secon

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Question 215375: Two cards are drawn in succession without replacement from a standard deck of 52 playing cards. What is the probability that the first card drawn was a diamond given that the second card drawn was not a diamond? (Round your answer to four decimal places.)
i cant seem to get this problem right as simple as it sounds!!!!!!
Found 2 solutions by drj, Edwin McCravy:
Answer by drj(1380) About Me  (Show Source):
You can put this solution on YOUR website!
Two cards are drawn in succession without replacement from a standard deck of 52 playing cards. What is the probability that the first card drawn was a diamond given that the second card drawn was not a diamond? (Round your answer to four decimal places.)

Step 1. Let P(A) be the probability of a diamond and P(B) is not a diamond or 1-P(A) for the first card. For the first card P(A)=13/52=0.25 and P(B)=1-P(A)=0.75

Step 2. Let Probability of a diamond occurring in the second card given that probaility of diamond occurring in the first card be given as P(A|B)= P(A and B)/P(B).

Step 3. The probability of a diamond in the second card is P(A and B)=13/51.

Step 4. P(A|B)=P(A and B)/P(B)= (13/51)/((39/52)=0.3399

Step 5. Answer is P(is a diamond on second card |not a diamond on first card)=P(A|B)=0.3399

I hope the above steps were helpful.

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Respectfully,
Dr J







Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
Two cards are drawn in succession without replacement from a standard deck of 52 playing cards. What is the probability that the first card drawn was a diamond given that the second card drawn was not a diamond? (Round your answer to four decimal places.)
i cant seem to get this problem right as simple as it sounds!!!!!!

There are 13 diamonds and 39 non-diamonds in a deck of 52.

The formula for a conditional probability is

P%28%22A%7CB%22%29+=+P%28A_and_B%29%2FP%28B%29

where %22A%7CB%22 means "A given B"

Let A be "the first card was a diamond"
Let B be "the second card was a non-diamond"

The probability on the left of the formula is a CONDITIONAL 
probability, but the two probabilities on the right are NOT.  
We are not to assume we are given anything when we calculate 
the right side.

"AND" indicates that we are to multiply and "OR" indicates
that we are to add.

P%28A_and_B%29+=+%2813%2F52%29%2839%2F51%29=+13%2F68



Going back to the conditional probability formula:

P%28%22A%7CB%22%29+=+P%28A_and_B%29%2FP%28B%29=%2813%2F68%29%2F%283%2F4%29=13%2F51

To ask me anything you don't understand about this problem, email
me at AnlytcPhil@aol.com, or ask me in your thank-you note.

Edwin