How to Solve this Question?

Question

Using a standard 52-card deck, find the following probabilities. Leave your answer as a reduced fraction.

c. If two cards are randomly selected, what is the probability of drawing any one suit first, followed by drawing a face card? 

The solution for solving it shows (10/52 * 12/51) + (3/52 * 11/51) = 3/52, but I'm not sure where they got these numbers for the numerator, such as with the 10 or 3. I assume possibly they could be the number of (ace - 10) and face amount (queen, king, and jack). But then wouldn't they be 40 and 12 since there are 4 suits? And I'm still unsure about the 10 and 11. Thank you so much for your time.

Answer 1

probability of a suit is 1/4. or 13/52. there are 4 suits, spaces, hearts, diamonds and clubs. 13 cards in each suit.

probability of a face card is 12/52 = 3/13. there are 3 face cards in each suit of 13 cards or 12 face cards in the deck of 52 cards.

1/4 x 3/13 = 3/(4)(13) = 3/52

or

13/52 x 12/52 = 156/2704 = 3/52

Hard to tell what the "solution" you have is about.

10 is from 10 non-face cards. 3 is from 3 face cards

12 comes from the suit cards left if the non face card was the same suit

51 is the remaining cards after one card is drawn

11? other than 11 makes it work out to the correct answer 3/52

11 seems to be how many cards left in the suit of 13 after removing 2 cards

a complicated way to get the answer. 1/4 x 3/13 is much easier.

they seem to be calculating getting a nonface card times getting the same suit as that nonface card

then adding what you'd get with a face card times getting the same suit