What's the probability of at least one zero appearing in a sequence of five digits?

Question

Possible digits range from 0 to 9. Explanation/work. Thank you !

Answer 1

Since there are nine digits other than zero, the probability of NO zeroes is (9/10)^5. So, the probability of at least one zero is 1 - (9/10)^5 = .40951.

Note: Since the question asks for "a sequence of five digits" as opposed to a number, I am assuming the sequence can start with leading zeroes.

Answer 2

I assume you can't have a leading zero. So the first digit can be 1 to 9. Second digit 0 to 9, etc.

Let's figure the chance of there *not* being a zero in the number at all.

The probability is 9/9 for the first digit, 9/10 for the second digit, 9/10 for the third digit, 9/10 for the fourth digit and 9/10 for the last digit.

Multiply this out and you get:
9/9 x 9/10 x 9/10 x 9/10 x 9/10 = 0.6561

But this is the probably that *no* zeroes appear. Therefore the probability that *at least one* zero appears is 1 minus this:

1 - 0.6561 = 0.3439

P(at least 1 zero, leading zeroes not allowed) ≈ 34.39%

Alternatively, if this is truly just a "sequence of five digits" and leading zeros *are* allowed, then you use this method, but modify it slightly.

The probability of no zeroes would be *9/10* x 9/10 x 9/10 x 9/10 x 9/10 = 0.5905

And the probability (assuming leading zeroes are allowed) of at least one zero would be 1 minus this.

1 - 0.5905 = 0.4095

P(at least 1 zero, leading zeroes allowed) ≈ 40.95%

Answer 3

P(at least 1) = 1 - P(none)

There are a total of 10*10*10*10*10 = 100000 ways to pick those 5 digits. Of these, 9*9*9*9*9 = 59049 ways of picking those 5 digits without using a 0.

P(none) = 59049/100000
P(at least 1) = 1 - 59049/100000
= 40951/100000
= 0.40951

Answer 4

One in fifty.

5 possible fields

A B C D E

Each field can have one of 10 numerals.

10 x 5 = 50

But that is for "at least one"