What is the probability of drawing any suit first, followed by drawing a face card?

Question

Using a standard 52 card deck, find the following probability:

if 2 cards are randomly selected, what is the probability of drawing any suit first, followed by drawing a face card? Assume without replacement.

Answer 1

Alan T's interpretation of the question (in his comment to Somya's post) is what is intended - based on getting it correct :) As he mentions, the problem is that the first card of a particular desired suit may or may not be a face card.

So these two mutually-exclusive cases have to be considered and their probabilites added together:

Case 1: P( suit & FC ) * P( FC | first was FC ) = (3/52)(11/51)

Case 2: P( suit & not FC ) * P( FC | first not FC) = (10/52)(12/51)

P(total) = (3*11+10*12)/(52*51) = 153/(52*51) = 9/(52*3) = 3/52 .

Interestingly, the probabilty is the same for the reverse situation: pick a face card and then pick a particular suit. In this case the 153 in the intermediate numerator is formed from 3*12+9*13 = 153.

Probably some deep principle at work here :)

Answer 2

I hope I understood the question correctly!

Since the first card is from the deck the probability it is from a suit is 1.

P(any suit, face card) = P(1st card is a face card. 2nd card is a face card) + P(1st card is not a face card, 2nd card is a face card)

= 12/52*11/52 + 40/52 * 12/52

= 0.226