Question

one ticket is selected at random from 100 tickets numbered 00,01,02, ....... 99
suppose X is sum and Y is the product of the digit found on the ticket

find the probability of x = 9 and y = 0

Answer 1

n ( s ) = 100

Y = { 01 , 02 ,03 ,04 ,05 , 06 ,07 , 08 , 09 , 10 , 20 ,30 , 40 ,50 , 60 ,70 ,80 , 90 }

so

n ( Y ) = 18

P ( Y ) = n ( Y ) / n ( S )

= 18 /100

= 9 /50

X = { 09 , 18 , 27 , 36 , 45 , 54 , 63 , 72 , 81 , 90 }

n ( X ) = 10

P ( X ) = n ( X ) / n ( S )

= 10 / 100

= 1 / 10

so probability of event X is 1/10 and event y is 9/50

Answer 2

hey your answer
Ransh sangwan

product of the digit found on the ticket = 0 , 
therefore one digit must be 0.
therefore possible outcomes = {00,01,02,03,04,05,06,07,08,09,10,20,30,40,50,60,70,80,90}
total number of outcomes = 19
and the sum of the digits = 9
since one digit is 0 , therefore other digit must be 9.
thus the possible number on the ticket is 09 or 90
number of favorable outcomes = 2
thus the required probability = 2/19