Question

The numbers 1 to 9 are randomly arranged to form a 9-digit number. what is the probability that the number obtained is divisible by 18?
(please write the solution)

Answer 1

Answer:

Hence the probability that the number is divisible by 18 is 4/9 or approx. 0.4444. or even 44.4%

Step-by-step explanation:

If we have 9 digits 1–9 placed randomly to form a 9-digit number without repetition and we want the probability of that number is divisible by 18, then we only need to look at the fact if it is divisible by 18, then it is also divisible by 2 and 9. Adding the digits 1–9 would give us the following (including checking for divisibility by 9):

(9 x 10)/2 =90/2 = 45

45÷9=5

By divisibility of 9, if the sum of the digits is divisible by 9, then the original number itself is divisible by 9. Since the number 45 is divisible by 9, we know that all the possible 9-digit numbers will be as well. The only thing we would have to look at is if it is even since the number has to be divisible by 2. 

There are 9 digits of which 4 are even (2, 4, 6 or 8).

So the probability that the number is divisible by 2 is 4/9.

Hence the probability that the number is divisible by 18 is 4/9 or approx. 0.4444.