Question

Suppose a 5-digit number is formed using the digits from 1 to 9 (without replacement). What is the probability that it will be an even number? ​

Answer 1

By finding the probability that the unit digit is even, we will see that the probability that the 5-digit number will be even is P = 4/9.

So we have a 5-digit number, let's say that it is:

abcde

Now, we have 5 "slots" where we can put numbers from 1 to 9 (without replacement).

We want to find the probability that the number will be even, so we only care for the last digit, e (it must be even).

Let's assume that the first digit at which we assign its value is this, e, the last one.

For this digit we have 9 options, {1, 2, 3, 4, 5, 6, 7, 8, 9}

4 of these options are even numbers.

So there are 4 out of 9 cases where the last digit is even, then the probability that the 5-digit number will be even is just:

P = 4/9

Now the other four digits can be anything (even or odd), so we only care for the probability above.

If you want to learn more about probability, you can read:

brainly.com/question/23044118