1988 AHSME Problems/Problem 12

Problem

Each integer $1$ through $9$ is written on a separate slip of paper and all nine slips are put into a hat. Jack picks one of these slips at random and puts it back. Then Jill picks a slip at random. Which digit is most likely to be the units digit of the sum of Jack's integer and Jill's integer?

$\textbf{(A)}\ 0\qquad \textbf{(B)}\ 1\qquad \textbf{(C)}\ 8\qquad \textbf{(D)}\ 9\qquad \textbf{(E)}\ \text{each digit is equally likely}$

Solution

We can draw a sample space diagram, and find that of the $9^2 = 81$ possibilities, $9$ of them give a sum of $0$, and each other sum mod $10$ (from $1$ to $9$) is given by $8$ of the possibilities (and indeed we can check that $9 + 8 \times 9 = 81$). Thus $0$ is the most likely, so the answer is $\boxed{\text{A}}$.


See also

1988 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 13
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