The total possible 4 digit pins from the each number between 0 and 9 is,
[tex]\begin{gathered} n(T)=10\cdot10\cdot10\cdot10 \\ =10000 \end{gathered}[/tex]The debit card has only one correct pin. So possibe outcome is,
[tex]n(P)=1[/tex]Determinen the probability for the figure out correct pin is,
[tex]\begin{gathered} P=\frac{n(P)}{n(T)} \\ =\frac{1}{10000} \\ =0.0001 \end{gathered}[/tex]Thus answer is 0.0001 or 1/10000.
