We are given that a secret number is a 4-digit integer created using the digits from 1 to 9. We are asked to find: a. The total number of possible secret numbers. b. The probability of selecting a secret number at random that is odd. c. The probability of selecting a secret number at random that has all four digits different.

1. Problem Description

We are given that a secret number is a 4-digit integer created using the digits from 1 to

9. We are asked to find:

a. The total number of possible secret numbers.

b. The probability of selecting a secret number at random that is odd.

c. The probability of selecting a secret number at random that has all four digits different.

2. Solution Steps

a. Finding the total number of possible secret numbers:

Since each of the four digits can be any of the 9 digits (1 to 9), and repetition is allowed, the total number of possible secret numbers is:

$9 \times 9 \times 9 \times 9 = 9^4$

b. Finding the probability of selecting an odd secret number:

For the secret number to be odd, the last digit must be odd. There are 5 odd digits between 1 and 9 (1, 3, 5, 7, 9). The first three digits can be any of the 9 digits. So, the number of odd secret numbers is:

$9 \times 9 \times 9 \times 5 = 9^3 \times 5$

The probability of selecting an odd secret number is the number of odd secret numbers divided by the total number of secret numbers:

$P(\text{odd}) = \frac{9^3 \times 5}{9^4} = \frac{5}{9}$

c. Finding the probability of selecting a secret number with all different digits:

For a secret number with all digits different, the first digit can be any of the 9 digits. The second digit can be any of the remaining 8 digits. The third digit can be any of the remaining 7 digits. The fourth digit can be any of the remaining 6 digits. So, the number of secret numbers with all digits different is:

$9 \times 8 \times 7 \times 6$

The probability of selecting a secret number with all digits different is the number of such secret numbers divided by the total number of secret numbers:

$P(\text{all different}) = \frac{9 \times 8 \times 7 \times 6}{9^4} = \frac{9 \times 8 \times 7 \times 6}{9 \times 9 \times 9 \times 9} = \frac{8 \times 7 \times 6}{9 \times 9 \times 9} = \frac{336}{729} = \frac{112}{243}$

3. Final Answer

a. The total number of possible secret numbers is $9^4 = 6561$.

b. The probability of selecting an odd secret number is $\frac{5}{9}$.

c. The probability of selecting a secret number with all different digits is $\frac{112}{243}$.