A bag contains 19 tickets, numbered from 1 to 19. A ticket is drawn and then another ticket is drawn without replacement. Find the probability that both tickets will show even numbers.

Let A be the event of drawing an even numbered ticket in first draw and B be the event of drawing an even numbered, ticket in the second draw. Then,

Required probability = P(A ∩ B) = P (A) P (B/A)  ….(i)

Since there are 19 tickets, numbered 1 to 19, in the bag out of which 9 are even numbered viz. 2, 4, 6, 8, 10, 12, 14, 16, 18. Therefore, P(A) = 9/19

Since the ticket drawn in the first draw is not replaced, therefore second ticket drawn is from the remaining 18 tickets, out of which 8 are even numbered.

∴ P (B/A) = 8/18 = 4/9

Hence, required probability = P (A ∩ B) = P(A)P(B/A) = 9/19 × 4/9 = 4/19