Question

A bag contains 25 tickets, numbered from 1 to 25. A ticket is drawn and then another ticket is drawn without replacement. Find the probability that both tickets will show even numbers.

Answer 1

There are 25 tickets numbered from 1 to 25, so the sample space is

S = {1, 2, 3, …., 25}, n(S) = 25

Number of even numbered tickets from 1 to 25 is

{2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24} = 12

Let A represents first ticket with even number and B represents second ticket with even number

Then the probability of both tickets being even number without replacement is

P(both tickets showing even number without replacement)

= P(A)P(B|A)

= \(\cfrac{12}{25}\times\cfrac{11}{24}\) (as there are 12 even numbered tickets out of 25 tickets in first draw, and 11 even numbered tickets out of 24 tickets in second draw as the tickets are not replaced)

Hence the required probability is \(\cfrac{11}{50}\)