# 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.

Answer: 3/50 = 0.06.

I hope you get the solution of “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.” This question had asked in SEBA Class 12 Board Exam and you can expect such type of rare question in upcoming exams. 