The probability would be 0.01458.
This is a binomial probability, since the probabilities are independent, there is a fixed number of trials, and there are two outcomes (either a 0 or not a 0). We use the formula:
[tex]_nC_r(p)^r(1-p)^{n-r}[/tex]
The probability of any of the digits being drawn is 1/10. Then we have:
[tex]_6C_3(0.1)^3(1-0.1)^{6-3}
\\
\\_6C_3(0.1)^3(0.9)^3
\\
\\\frac{6!}{3!3!}(0.1)^3(0.9)^3
\\
\\20(0.1)^3(0.9)^3 = 0.01458[/tex]
