+195 The three-digit positive integer has a ones digit of 3. What is the probability that is divisible by 3? Express your answer as a common fraction.
CPhill: The question says "The three-digit positive integer has a ones digit of 3". Doesn't that mean that the 3rd digit is fixed at 3? E.G. 103, 113, 123, 133, 143, 153 163, 173, 183, 193......and so on for a total of 9 x 10 = 90 3-digit numbers ALL ending in 3. And there would be: 90/3 =30 numbers out the 90 divisible by 3. Or 30/90 =1/3.
+130071
For this to be divisible by 3, the sum of the first two digits must be divisible by 3
There will be (99-12)/ 3 + 1 = 30 such numbers
And there are (999-100) + 1 = 900 three digit numbers....so
The probability that one ends in "3" and is divisible by 3 is just
30 / 900 = 1/30
CPhill: The question says "The three-digit positive integer has a ones digit of 3". Doesn't that mean that the 3rd digit is fixed at 3? E.G. 103, 113, 123, 133, 143, 153 163, 173, 183, 193......and so on for a total of 9 x 10 = 90 3-digit numbers ALL ending in 3. And there would be: 90/3 =30 numbers out the 90 divisible by 3. Or 30/90 =1/3.
