100 tickets are numbered as `00, 01, 02).... 0910, 11, 12, ..... 99` out of them one ticket is drawn at random. The probability that the sum of the digits of the number on the ticket is 9 is

100 tickets are numbered as 00,01,02,.......,09,10,11,.......99. When a ticket is drawn at random from them and if B is the event of getting 0 as the product of the numbers on the ticket, then P(B)=

100 tickets are numbered as 00, 01, 02, ……..09, 10, 11, ….99. When a ticket is drawn at random from then and if A is the event of getting 9 as the sum of the numbers on the ticket, then P(A) =

100 tickets are numbered as 00 ,01, 02, ……09, 10, 11….99. When a ticket is drawn at random from them and if b is the event of getting 0 as the product of the numbers on the ticket, then P(b) =

One hundred tickets are numbered as 00,01,02,…..,09,10,11,….,99 and one ticket is drawn at random from them. If A is the event of getting 9 as the sum of the numbers on the ticket and B is the event of getting 0 as the product of the numbers on the tickets then P(AnnB)=

Four tickets numbered 00,01,10,11 are placed in a bag.A ticket is drawn at random and replaced.Again a ticket is drawn at random.The probability that the sum of the numbers on the tickets drawn is 21

Out of 7 tickets consecutively numbered 3 are drawn at random. The probability for the numbers on the tickets to be in A.P. is

Out of 7 tickets consecutively numbered 3 are drawn at random. The probability for the numbers on the tickets to be in A.P. is

Out of 21 tickets numbered 10, 11, 12,…, 30, three tickets are drawn at random. Find the probability that the numbers on these tickets are in A.P.

A bag contains tickets numbered 1 to 20 . Two tickets are drawn at random , Find the probability that sum of the two numbers on the tickets is even .