# Doubt from probability

Take cases

Fix say,5,then in the next 4 boxes, permute available numbers except 3,6 and 9

Then fix,say,3 then permute available numbers except 5,6 and 9

Then fix 6 and permute available numbers except 5,6 and 3

Do the same for 9

And then make a case which includes two or more than 2 numbers from the set {3,6,9} in the targeted 5 number box and combine all these and permute them with numbers outside this box considering this box to be one unit and divide by the total possibilities

Or an easier way would be to find those numbers which aren't divisible by 5 or 3 in a similar manner as I stated above and subtract that from 1

This'll be a faster method

*Subtract the probability of the occurrence of those cases from 1

I am trying to solve and in this context i want to share my opinion.Pls see my approach and share your views.It is not solved yet.

Put 5 in place of X5 then condition is fulfilled.so p(N5)=1/9

I have wrongly considered p(N15)=0 but it is not zero

Sir i will confirm the answer once again . I think the answer is wrong thanks for your solution