Probability of Dialing Correct Digits 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. 
 A: Some careful thought will show you why the total number of possibilities for the last two digits is not 18. Suppose we fix the 6th number to be 9, for example. Then there are 9 two-digit combinations that could be dialed: (9,1), (9,2), (9,3) ... (9,9). But suppose the 6th digit was not 9, but instead 8. We again have another 9 possibilities we could dial. How about fixing the 6th at 7? Another 9 possibilities. Proceeding in this way, we see that the total number of possibilities for the last two digits is 9 * 9 = 81. Now, since only 1 combination of two numbers in the correct order is the correct combination, the probability of choosing the correct combination is 1/81.
Another way to think about it is this: dialing the 6th number correctly and 7th number correctly are two statistically independent events. Since there are 9 possible digits that could be entered for both the 6th and 7th number, but only 1 correct number for each, we have a 1/9 chance of getting each number right. Since they are independent, we can multiply them together to get the joint probability of getting both right: 1/9 * 1/9 = 1/81.
A: 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?
