Question

You randomly select two cards from a standard 52-card deck. What is the probability that the first card is not a face card (a king, queen, or jack) and the second card is a face card if you replace the first card before selecting the second?

Answer 1

There are 3 face cards per suit, so 12 face card in total. This implies that there are 40 non-face cards in the deck. So, the probablity of picking a face card is 12/52, and the probability of picking a non-face card is 40/52.

Since we replace the first card, the two picks follow the same probability, because we're picking from a whole deck every time.

The probability of two independent events happening is the product of the two probabilities, so we have

$\dfrac{12}{52}\cdot\dfrac{40}{52}=\dfrac{30}{169}$

Answer 2

Answer:

Step-by-step explanation:

I'll give this one a shot......we have the following combos as possibilites

(2,10)  (3,9) (4,8) (5,7) (6,6)

In the first four cases we have 16 ways to form each .....choose any one of four cards from the first given rank and choose the same from the second one. So C(4,1) * C(4,1) = 16

In the last case we just want to choose any two of the four "6s" and there are C(4.2) ways to do that = 6 ways

So, the total possibilities  = 16*4 + 6 = 70

And the total possibilities from choosing any 2 cards from 52 = C(52, 2)   = 1326

So.....  70 / 1326 =  about .053 = about 5.3%

Someone else needs to check this, since I'm never sure about these "counting" problems  !!!