Question

How many integers are there from 1,000 through 9,999? (b) How many odd integers are there from 1,000 through 9,999? (c) How many integers from 1,000 through 9,999 have distinct digits? (d) How many odd integers from 1,000 through 9,999 have distinct digits? (e) What is the probability that a randomly chosen four-digit integer has distinct digits? What is the probability that a randomly chosen four-digit integer has distinct digits and is odd?

Answer 1

Answer:

(a) $Total= 9000$

(b) $Total = 4500$

(c) $Total = 4536$

(d) $Total = 2240$

(e) $Pr = 0.5040$ --- Probability of distinct digits

$Pr = 0.2489$ --- Probability of odd distinct digits

Step-by-step explanation:

Solving (a): Integers from 1000 to 9999

To do this, we simply add 1 to the range.

i.e.

$Total= Range +1$

Range is the difference between the given interval.

$Total= 9999 - 1000 +1$

$Total= 9000$

Solving (b): Odd integers

This implies that the last digit must be any of 1, 3, 5, 7 and 9 (i.e. 5 digits)

The first digit can not be 0 (i.e any of the remaining 9 digits)

There is no restriction to other digits

Numbers from 1000 to 9999 are 4 digits, so the possible selection are:

$First= 9\\Second = 10\\Third = 10\\Last = 5$

Total selection is:

$Total = 9 * 10 * 10 * 5$

$Total = 4500$

Solving (c): Distinct digits

This implies that all 4 digits are different and the first can not be 0.

So, we have:  

$First = 9$ i.e. (1 - 9)

$Second = 9$ i.e. (0 - 9) minus the first digit

$Third = 8$

$Fourth = 7$

Total selection is:

$Total = 9*9*8*7$

$Total = 4536$

Solving (d): Odd digits that are distinct

This implies that the last digit must be any of 1, 3, 5, 7 and 9 (i.e. 5 digits)

The first digit can not be 0 and must be different from the last (i.e 8 digits)

The second digit must be different from the first and the last(i.e. 8 digits)

The third digit must be different from the three other digits (i.e. 7 digits)

So,

$Total = 5 * 8 * 8 * 7$

$Total = 2240$

Solving (e): Probability that a number is distinct

In (a), total possible digits is 9000

In (c), total distints are 4536

So, the probability is:

$Pr = \frac{4536}{9000}$

$Pr = 0.5040$

In (d), total odd distinct digits are 2240

So, the probability is:

$Pr = \frac{2240}{9000}$

$Pr = 0.2489$