Probability of Finding Disconnected Patterns in Digit Positions of Random Integer Sequence

Question

Let's say I have a random sequence of 9 digits.

x x x x x x x x x

What is the probability of finding '9' in the first position and '93' in the fourth and fifth positions?

9 x x 9 3 x x x x

Now suppose we allow ranges of numbers as patterns. With the same 9-digit sequence, what is the probability of finding '9' in the first position and a number in the range (70 - 88) or (90 - 92) or (94 - 99) in the fourth and fifth positions?

9 x x 70 x x x x
9 x x 71 x x x x
    ...
9 x x 88 x x x x

OR

9 x x 90 x x x x
9 x x 91 x x x x
9 x x 92 x x x x

OR

9 x x 94 x x x x
9 x x 95 x x x x
    ...
9 x x 99 x x x x

Answer 1

The total number of sequences is $10^{9}.$

Use constructive counting. If $9$ is in the first position and $93$ takes up positions three and four, we now have only $10^{6}$ choices (we select the remaining $6$ digits).

Our answer is thus $\frac{10^{6}}{10^{9}} = \boxed{\frac{1}{1000}}.$

Answer 2

For the second part of your question, there are $10$ possibilities for the first position, of which $1$ qualifies, for a probability of $1/10$. There are $100$ possibilities for the fourth and fifth positions, of which $19+3+6 = 28$ qualify, for a probability of $18/100 = 9/50$. Thus, the probability that both qualify is

$$ \frac{1}{10} \times \frac{9}{50} = \frac{9}{500} $$