# A bag contains 20

There are 20 tickets numbered from 1 to 25, so the sample space is

S = {1, 2, 3, …., 20}, n(S) = 20

Number of even numbered tickets from 1 to 20 is

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

Number of odd numbered tickets from 1 to 20 is

{1, 3, 5, 7, 9, 11, 13, 15, 17, 19} = 10

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

Then the probability of first ticket having even number and second ticket having odd number without replacement is

(as there are 10 even numbered tickets out of 20 tickets in first draw, and 10 odd numbered tickets out of 19 tickets in second draw as the tickets are not replaced)

Hence the required probability is

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