Question

Consider all six-digit numbers that can be created from the digits 0-9 where the first and last digits must be odd and no digit can repeat. What is the probability of choosing a random number that starts with 9 from this group? Enter a fraction or round your answer to 4 decimal places, if necessary

Answer 1

Using the Fundamental Counting Theorem, it is found that there is a 0.2 = 20% probability of choosing a random number that starts with 9 from this group.

What is the Fundamental Counting Theorem?

It is a theorem that states that if there are n things, each with $n_1, n_2, \cdots, n_n$ ways to be done, each thing independent of the other, the number of ways they can be done is:

$N = n_1 \times n_2 \times \cdots \times n_n$

In this problem, we have that:

• The first and last digit have to be odd, hence $n_1 = 5, n_6 = 4$.

• The others cannot repeat, hence: $n_2 = 8, n_3 = 7, n_4 = 6, n_5 = 5$.

Then, the total number is:

N = 5 x 4 x 8 x 7 x 6 x 5 = 33600.

Starting with 9, we have that:

• The first digit is 9, the last digit has to be an odd which is not 9, hence $n_1 = 1, n_6 = 4$.

• The others cannot repeat, hence: $n_2 = 8, n_3 = 7, n_4 = 6, n_5 = 5$.

Then:

N9 = 1 x 4 x 8 x 7 x 6 x 5 = 6720.

A probability is given by the number of desired outcomes divided by the number of total outcomes, hence:

p = N9/N = 6720/33600 = 0.2.

0.2 = 20% probability of choosing a random number that starts with 9 from this group.

To learn more about the Fundamental Counting Theorem, you can check https://brainly.com/question/24314866