# In a 3×3 matrix, entries aij are selected randomly from the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 with replacement where each element aij is a three digit number. Find the probability that each element in each row is divisible by 15.

Question: In a 3×3 matrix, entries aij are selected randomly from the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 with replacement where each element aij is a three-digit number. Find the probability that each element in each row is divisible by 15.

Solution:

Given: Each aij is a three-digit number
So, the hundredth place can’t be zero (0).

The total number of each dement of arrangement of 3 digits no. is 9 X 10 X 10 =900

Also Given: Each element in each now is divisible by 15. That means each element is divisible by both 3 and 5.

We know that:

• The Numbers divisible 5 contain either 0 or 5 in the unit digit place.
• For divisibility of 3, the sum of digits of the numbers must be divisible by 3.

Now every element in each row is divisible by 15 must end with 0 and 5 also it must satisfy the criteria of the sum of each digit must be divisible by 3.

Case 1: When the unit digit is 0

S1={ { (1,2,0), (1,5,0), (1,8,0), (2,1,0), (2,4,0), (2,7,0), (3,0,0), (3,3,0), (3,6,0)…(9,0,0), (9,3.0), (9,6,0)}

No. of possibility is, 3×9 = 27

Case 2: When unit digit is 5.

S2 = {(1,0,5), (1,3,5), (1,8, 5), (2,2,5)… }

Number of possibilities: 27
Total no of possibility: n(S1) + n(S1) = 27 + 27 = 54

Probability = (3 digit no. divisible by 15) / (Total 3 digit no,)

So, Probability =  54/900 = 3/50.

Therefore, the probability that each element in each row is divisible by 15 is 3/50.