All four aces and all twelve face cards are taken out of a standard pack of 52 playing cards. One card is then drawn at random from what remains. What is the probability that it is a black card showing a multiple of 3?

Removing 4 + 12 = 16 cards leaves 36 — the cards 2 to 10 in each of the four suits. Among 2 to 10 the multiples of 3 are 3, 6 and 9, and 'black' keeps only spades and clubs, so $3\times 2=6$ cards qualify and the probability is 6/36 = 1/6. Dropping the word 'black' gives 12/36 = 1/3; leaving the denominator at the original 52 gives 6/52 = 3/26, the commonest slip once cards have been removed; and 1/2 is just the chance of a black card.