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.

A) 4/9     B) 3/9    C) 2/9     D) 1/9 

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)

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)

=9/19x4/9=4/9