Question

a box contains 36 tickets numberred from 1 to 36 one ticket drawn random find the probably that the number on the ticket is either divisible by 3 or is a perfect square

Answer 1

The numbers which satisfy your condition are :
3,6,9,12,15,18,21,24,27,30,33,36
&
1,4,9,16,25,36
there are 16 of these numbers
(9,36 are repeated)
probability = 16/36
= 0.44444

Answer 2

n(S)= 36
Let A be the event that the number is divisible by 3.
A = { 3,6,9,12,15,18,21,24,27,30,33,36 }
n(A) = 12 P(A) = 12/36
Let B be the event that the number on the ticket is a perfect square.
P(B) = { 1,4,9,16,25,36 }
n(B) = 6
P(B) = 6/36
Now AB = { 9, 36}
n( A B )= 2 
P( A B ) = 2/36
The required probability is P(A U B)
P(A U B) = P(A) + P(B) - P( AB)
= 12/36 + 6/36 -2/36
= 4/9