-1

An integer from $100$ through $999$ inclusive is to be chosen at random. What is the probability that the number chosen will have $0$ as at least one digit? Here is a list of answers

user8725
  • 3
  • 2
    One solution: count the number of integers [in that range] that have at least one zero [list them all out?], and divide by 900. You can be a little more clever with the counting to save time. – angryavian Jun 09 '16 at 23:37
  • How many integers from $100$ to $199$ contain zeroes in their representation? – Joffan Jun 09 '16 at 23:42
  • lol body_ACT_1112_-_51 – Mark C Oct 25 '17 at 17:35

3 Answers3

2

There are $999-99=900$ numbers in the range from $100$ to $999$ inclusive. The number of three-digit numbers using only digits $1$ through $9$ inclusive is $9\times9\times9=729.$ Thus the probability that a random three-digit number has no zeroes in its decimal representation is $$\frac{9\times9\times9}{900}=\frac{9\times9}{100}=81\%$$ and so the probability that it has at least one zero is $$100\%-81\%=19\%$$ which is the same as $$\frac{171}{900}=\frac{19}{100}=19\%.$$

bof
  • 78,265
1

There are $90=10\times9$ numbers which have a 0 as the last digit, i.e. of the form _ _ $0$.

There are $90=10\times9$ numbers which have a 0 as the second digit, i.e. of the form _ $0$ _.

However in counting these those with a $0$ as both second and last digit have been counted twice. There are 9 numbers which have a $0$ as both second and last digit, i.e. of the form _ $0$ $0$. These $9$ must be subtracted off as they have been counted twice.

So the answer is $90+90-9=171$.

Ian Miller
  • 11,844
1

You can generate your number by independently picking the first digit uniformly from $\{1,\ldots,9\}$ and picking the second and third digits uniformly from $\{0,\ldots,9\}$. The first digit will not be a zero. Let $A$ be the event that the second digit is zero, and $B$ the event that the third digit is zero. You want $P(A\cup B)$ which equals $P(A)+P(B)-P(A\cap B)$. Argue that $P(A)=P(B)=\frac1{10}$ and $P(A\cap B)=\frac1{100}$, so the answer is $\frac1{10}+\frac1{10}-\frac1{100}=\frac{19}{100}$.

grand_chat
  • 38,951