Consider all four-digit numbers that can be made from the digits 0-9 (assume that numbers cannot start with 0). What is the probability of choosing a random number from this group that is less than or equal to 5000

0.44

Step-by-step explanation:

First, let's count the total number of four-digit numbers.

Let's think it as four empty slots:

_ _ _ _

In each slot, we can put a single digit.

In the first slot, we have 9 options {1, 2, 3, 4, 5, 6, 7, 8, 9}

The zero is not an option here, because if there was a 0, this would be a 3 digit number, then the 0 can not be in the first slot.

For the second slot we have 10 options {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}

for the third slot we have 10 options {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}

for the fourth slot we have 10 options {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}

The total number of combinations is equal to the product between the numbers of options for each slot, this means that we have:

Combinations = 9*10*10*10 = 9000

So there are 9000 different 4 digit numbers.

Now we want to find the probability of selecting a number at random that is equal or smaller than 5000.

Let's see the number of 4-digit numbers that are equal or smaller than 5000.

The smallest 4-digit number is 1000.

Then the number of numbers between 1000 and 5000 is:

5000 - 1000 = 4000

This means that there are 4000 4-digit numbers that are equal or smaller than 5000.

Now, the probability of selecting one of them at random is equal to the quotient between the number of 4-digit numbers that meet the condition (4000) and the total number of 4-digits (9000)

Then te probability is:

P = 4000/9000 = 4/9 = 0.44

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i believe the answer is d.

hope it you! : d