Question

I) A box contains 36 tickets numbered from 1 to 36. One ticket
is drawn at random. Find the probability that the number
on the ticket is either divisible by 3 or 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

Answer:17/36

Step-by-step explanation:Total no. of outcomes =36

no.s divisible by 3 are (3,6,9,12,15,118,21,24,27,30,33,36)=12

no.s that are perfect square are (1,4,9,16,25,36)=6

therefore no. of favorable outcomes=12+5(not 6, as the no. 36 is already counted in one)=17

so, the the probability that the number on the ticket is either divisible by 3 or a perfect square is  17/36