2008 iTest Problems/Problem 44

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Problem

Now Wendy wanders over and joins Dr. Lisi and her younger siblings. Thinking she knows everything there is about how to work with arithmetic series, she nearly turns right around to walk back home when Dr. Lisi poses a more challenging problem. "Suppose I select two distinct terms at random from the $2008$ term sequence. What's the probability that their product is positive?" If $a$ and $b$ are relatively prime positive integers such that $a/b$ is the probability that the product of the two terms is positive, find the value of $a+b$.

Note: Dr. Lisi’s sequence is $-1776, -1765, -1754 \cdots$

Solution

The common difference of the arithmetic sequence is $11$. To find the number of terms that are negative (and the number of terms that are positive), we can write (and solve) an inequality.

\begin{align*} -1776 + 11(n-1) &< 0 \\ 11(n-1) &< 1776 \\ n-1 &< 161\frac{5}{11} \\ n &< 162\frac{5}{11} \end{align*}

Thus, there are $162$ negative terms and $1846$ positive terms. We can use complementary counting to calculate the probability of getting a negative product because it is easier than adding the probabilities where both numbers are either positive or negative.


There are $\binom{2008}{2}$ ways to choose two numbers from the sequence, and there are $162 \cdot 1846$ ways to choose one positive and one negative number. The probability of having a negative product is $\frac{162 \cdot 1846}{1004 \cdot 2007} = \frac{8307}{55973}$, so the probability of having a positive product is $1 - \frac{8307}{55973} = \frac{47666}{55973}$. Therefore, $a+b = \boxed{103639}$.

See Also

2008 iTest (Problems)
Preceded by:
Problem 43
Followed by:
Problem 45
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