Assuming that the number 7 must appear at least once, we find its complement first, which is not a phone number with number 7. The probability of not having a 7 is 9/10 (there are 10 possible digits). Hence,

\(\displaystyle{P}{\left({n}{o}{7}\right)}={\left(\frac{{9}}{{10}}\right)}\cdot{\left(\frac{{9}}{{10}}\right)}\cdot{\left(\frac{{9}}{{10}}\right)}\cdot{\left(\frac{{9}}{{10}}\right)}\cdot{\left(\frac{{9}}{{10}}\right)}\cdot{\left(\frac{{9}}{{10}}\right)}\cdot{\left(\frac{{9}}{{10}}\right)}={\left(\frac{{9}}{{10}}\right)}^{{7}}\)

Hence, the probability that there is at least 1 number 7 is:

P(at least one 7)=1-P(no 7)

P(at least one 7)=\(\displaystyle{1}-{\left(\frac{{9}}{{10}}\right)}^{{7}}\)

P(at least one 7)\(\displaystyle\sim{0.522}\to{52.2}\%\)

\(\displaystyle{P}{\left({n}{o}{7}\right)}={\left(\frac{{9}}{{10}}\right)}\cdot{\left(\frac{{9}}{{10}}\right)}\cdot{\left(\frac{{9}}{{10}}\right)}\cdot{\left(\frac{{9}}{{10}}\right)}\cdot{\left(\frac{{9}}{{10}}\right)}\cdot{\left(\frac{{9}}{{10}}\right)}\cdot{\left(\frac{{9}}{{10}}\right)}={\left(\frac{{9}}{{10}}\right)}^{{7}}\)

Hence, the probability that there is at least 1 number 7 is:

P(at least one 7)=1-P(no 7)

P(at least one 7)=\(\displaystyle{1}-{\left(\frac{{9}}{{10}}\right)}^{{7}}\)

P(at least one 7)\(\displaystyle\sim{0.522}\to{52.2}\%\)