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) =

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)=

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

if 3tickets are drawn at random from 41 tickets,numbered 1 to 41 find probability that the numbers on the tickets form a G.P.

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

Four tickets marked 00, 01, 10 , 11 respectively are placed in a bag. A ticket is drawn at random five times being replaced each time. The probability that the sum of the numbers on the tickets is 22 is

A ticket is drawn at random from 30 tickets numbered from 1 to 30. Find the probability that the number on the ticket drawn is divisible by 5 or 6.

Four tickets are marked 00, 01, 10 and 11 respectively, are placed in a bag. A ticket is drawn at random five times, being replaced each time. The probability that the sum of the numbers on the five tickets drawn is 24, is