A single card is drawn from a standard deck. Find the probability of the following event. (Enter your probability as a fraction.) Drawing a black card or a face card

Answer 1

If you think about this, there are 26 black cards and an additional K, Q, and J of hearts and of diamonds. That's 26 plus 6 or 32 cards. So the probability will be 36/52 (which can be reduced to 9/13).

The "math" way to do this problem is:

Let A represent black cards and B represent face cards

P(A or B) = P(A) + P(B) - P(A and B)

The P(A) = 26/52 (there are 26 total black cards)

The P(B) = 12/52 (there are 12 total face cards - a K, Q. J in each of 4 suits)

The P(A and B) = 6/52 (There are 6 black face cards)

So P(A or B) = P(A) + P(B) - P(A and B) = 26/52 + 12/52 - 6/52 = 32/52 = 9/13

All this is saying is to add the probabilities of the two events together but don't "double count". When you counted the black cards (26 of them), you counted the black face cards and when you counted just the face cards (12 of them) you also counted the black face cards. You can't (don't get to) count them twice so subtract out the double count.

Answer 2

If you think about this, there are 26 black cards and an additional K, Q, and J of hearts and of diamonds. That's 26 plus 6 or 32 cards. So the probability will be 36/52 (which can be reduced to 9/13).

The "math" way to do this problem is:

Let A represent black cards and B represent face cards

P(A or B) = P(A) + P(B) - P(A and B)

The P(A) = 26/52 (there are 26 total black cards)

The P(B) = 12/52 (there are 12 total face cards - a K, Q. J in each of 4 suits)

The P(A and B) = 6/52 (There are 6 black face cards)

So P(A or B) = P(A) + P(B) - P(A and B) = 26/52 + 12/52 - 6/52 = 32/52 = 9/13

All this is saying is to add the probabilities of the two events together but don't "double count". When you counted the black cards (26 of them), you counted the black face cards and when you counted just the face cards (12 of them) you also counted the black face cards. You can't (don't get to) count them twice so subtract out the double count.

Answer 3

If you think about this, there are 26 black cards and an additional K, Q, and J of hearts and of diamonds. That's 26 plus 6 or 32 cards. So the probability will be 36/52 (which can be reduced to 9/13).

The "math" way to do this problem is:

Let A represent black cards and B represent face cards

P(A or B) = P(A) + P(B) - P(A and B)

The P(A) = 26/52 (there are 26 total black cards)

The P(B) = 12/52 (there are 12 total face cards - a K, Q. J in each of 4 suits)

The P(A and B) = 6/52 (There are 6 black face cards)

So P(A or B) = P(A) + P(B) - P(A and B) = 26/52 + 12/52 - 6/52 = 32/52 = 9/13

All this is saying is to add the probabilities of the two events together but don't "double count". When you counted the black cards (26 of them), you counted the black face cards and when you counted just the face cards (12 of them) you also counted the black face cards. You can't (don't get to) count them twice so subtract out the double count.

Answer 4

If you think about this, there are 26 black cards and an additional K, Q, and J of hearts and of diamonds. That's 26 plus 6 or 32 cards. So the probability will be 36/52 (which can be reduced to 9/13).

The "math" way to do this problem is:

Let A represent black cards and B represent face cards

P(A or B) = P(A) + P(B) - P(A and B)

The P(A) = 26/52 (there are 26 total black cards)

The P(B) = 12/52 (there are 12 total face cards - a K, Q. J in each of 4 suits)

The P(A and B) = 6/52 (There are 6 black face cards)

So P(A or B) = P(A) + P(B) - P(A and B) = 26/52 + 12/52 - 6/52 = 32/52 = 8/13

All this is saying is to add the probabilities of the two events together but don't "double count". When you counted the black cards (26 of them), you counted the black face cards and when you counted just the face cards (12 of them) you also counted the black face cards. You can't (don't get to) count them twice so subtract out the double count.