A 3 -digit number is chosen at random. Given that it is a multiple of 4 , the probability that it has exactly two even digits is

Step-by-step solution

The total number of 3 -digit numbers that are multiples of 4 is 996 - 100 4 + 1 = 225 . Any multiple of 4 must end in an even digit. For the number to have exactly two even digits, the first two digits must contain exactly one even digit and one odd digit. This gives two cases: Even-Odd-Even and Odd-Even-Even. Case 1: Even-Odd-Even The last two digits must form a multiple of 4 . If the tens digit is odd ( 1, 3, 5, 7, 9 ), the units digit must be 2 or 6 . This gives 5 2 = 10 pairs for the last two digits. The first