A card is drawn at random from a deck of ordinary playing cards. What is probability that it is a face, a king or a diamond?

A card is drawn at random from a deck of ordinary playing cards. What is probability that it is a face, a king or a diamond?

There are 52 possible outcomes Hence, 52 is the denominator of probability fraction. Can I say that there are 13 diamonds plus 3 kings that are not diamonds for a total of 16..?

• I think it is a good idea to have a look at the answers of your previous question before posting a new question: math.stackexchange.com/questions/1507424/… – callculus Nov 1 '15 at 7:00
• Possibly. I don't quite understand the face part. If this includes, as usual, Jack and Queen, there are $13$ diamonds and $9$ non-diamond face cards. – André Nicolas Nov 1 '15 at 7:01
• calculus- what you want to say...? – bia Nov 1 '15 at 7:12
• @calculus-he question was asked by the same author.... – tatan Nov 1 '15 at 7:14
• @bia Sorry I didn´t recognize that you have posted comments. – callculus Nov 1 '15 at 7:16

A king is also a face card, so favorable outcomes $= 12$ face cards $+ 10$ "non-face" diamonds.

• can you please elaborate...i don't get it... – bia Nov 2 '15 at 5:11
• There are $3$ face cards, $J-Q-K$ in each suit. You have just to ensure that you don't double count. The King is included in the face cards, so we don't count it again. Also, since we considered all 12 face cards, we can't again include the 3 face cards of diamonds, so have counted 13-3 = 10 diamonds. $Pr = \frac{22}{52} = \frac{11}{26}$ – true blue anil Nov 2 '15 at 6:51

The total cards in an ordinary deck of cards is 52. Excluding the joker. the probability is given by = (total number of favourable outcomes/Total outcomes)

1. Face=12/52 (there are 3 face cards in each group therefore multiply by 4)

2. King=4/52 (there is only one king card in one group therefore multiply by4)

3. Diamond = 13/52 (there are 13 cards of diamond)

• Be careful. Note that the kings are face cards and that three of the diamonds are also face cards. – N. F. Taussig Nov 1 '15 at 9:25