Total number of all favorable cases is n(S) = 52
Let A be the event that first card drawn is a king. There are four kings in the pack. Hence, the probability of the first card is a king is
P(A) = \(\cfrac{4}{52}\)
Let B be the event that second card is an ace without replacement. Then there are 4 aces in the pack as the cards are not replaced. Therefore, the probability of the second card is an ace is
P(B|A) = \(\cfrac{4}{51}\)
Then the probability of getting first is a king and the second is an ace without replacement is
= P(A)P(B|A)
⇒ \(\cfrac{4}{52}\times\cfrac{4}{51}\) (as there are 4 kings out of 52 cards in first draw, and 4 aces out of 51 cards in the second draw as the cards are not replaced)
⇒ \(\cfrac{4}{663}\)
The probability that first is a king and the second is an ace without replacement is \(\cfrac{4}{663}\)
