Question

if one three - digit number (0 cannot be a left digit) is chosen at random from all those that can be made from the following set of digits, find the probability that the one chosen is not a multiple of 5.
{0, 1, 2, 3, 4, 5, 6, 7, 8}
the probability that the one chosen is not a multiple of 5 is

Explanation:

Step1: Calculate total three - digit numbers

The first digit (hundreds place) can be chosen in 8 ways (1 - 8). The second digit (tens place) can be chosen in 9 ways (0 - 8). The third digit (units place) can be chosen in 9 ways (0 - 8). So the total number of three - digit numbers is $8\times9\times9=648$.

Step2: Calculate three - digit numbers that are multiples of 5

Case 1: Units digit is 0. The first digit (hundreds place) can be chosen in 8 ways (1 - 8) and the second digit (tens place) can be chosen in 9 ways (0 - 8). So there are $8\times9 = 72$ such numbers.
Case 2: Units digit is 5. The first digit (hundreds place) can be chosen in 8 ways (1 - 8) and the second digit (tens place) can be chosen in 9 ways (0 - 8). So there are $8\times9=72$ such numbers.
The total number of three - digit numbers that are multiples of 5 is $72 + 72=144$.

Step3: Calculate three - digit numbers that are not multiples of 5

The number of three - digit numbers that are not multiples of 5 is $648-144 = 504$.

Step4: Calculate the probability

The probability $P$ that a chosen number is not a multiple of 5 is $\frac{504}{648}=\frac{7}{9}$.

Answer:

$\frac{7}{9}$