Jason remembers only the first five digits of a seven-digit phone number, but he is sure that neither of the last two digits is zero. If he dials the first five digits, and then dials two more digits, each chosen at random from the nonzero digits, what is the probability that he will dial the correct number?
Why isn't the total number of possibilities for the last two digits $18$?
I get $18$ by adding the number of digits possible for the first and second unknown terms.
Why isn't the possibilities of choosing the correct two digits $2/18$?
I get $2$ because i think the first and second digits have a $1+1=2$ correct chance.
Best Answer
How many cells does the following table have? $$ \begin{array}{c|c|} &1&2&3&4&5&6&7&8&9\\ \hline 1\\ \hline 2\\ \hline 3\\ \hline 4\\ \hline 5\\ \hline 6\\ \hline 7\\ \hline 8\\ \hline 9\\ \hline \end{array} $$ And if row numbers denote the second last digit and column numbers the last digit, how many choices do we then have for the last two digits?
A last question: what is $9\times 9$ equal to?