Question

(a) how many integers are there from 1,000 through 9,999?
9,000
(b) how many odd integers are there from 1,000 through 9,999?
4,500
(c) how many integers from 1,000 through 9,999 have distinct digits?
4,536
(d) how many odd integers from 1,000 through 9,999 have distinct digits?
2,240
(e) what is the probability that a randomly chosen four - digit integer has distinct digits?
0.504
what is the probability that a randomly chosen four - digit integer has distinct digits and is odd?
0.4938
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Explanation:

Step1: Determine the total number of four - digit integers

The total number of integers from \(1000\) through \(9999\) is \(9999 - 1000+1=9000\) (using the formula \(n = b - a + 1\) where \(a = 1000\), \(b = 9999\)).

Step2: Determine the number of odd four - digit integers with distinct digits

For an odd four - digit integer with distinct digits:

• The last digit (units place) can be chosen in \(5\) ways (\(1,3,5,7,9\)).
• The first digit (thousands place) can be chosen in \(8\) ways (it can't be \(0\) and can't be the digit chosen for the units place).
• The second digit (hundreds place) can be chosen in \(8\) ways (it can't be the digit chosen for the thousands and units place).
• The third digit (tens place) can be chosen in \(7\) ways (it can't be the digits chosen for the thousands, hundreds and units place).

By the multiplication principle, the number of odd four - digit integers with distinct digits is \(8\times8\times7\times5=2240\).

Step3: Calculate the probability

The probability \(P\) that a randomly chosen four - digit integer has distinct digits and is odd is given by the formula \(P=\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}\).
The number of favorable outcomes (odd four - digit integers with distinct digits) is \(2240\) and the total number of four - digit integers is \(9000\).
So \(P=\frac{2240}{9000}=\frac{224}{900}=\frac{56}{225}\approx0.2489\)

Answer:

\(0.2489\)