ExplanationThe trap answer in this problem is \(\frac{1}{6}\times \frac{1}{6}\).
This is not the answer to the question being asked—rather, this is the answer to the question, “What is the probability of picking a red chip and then a blue chip that both have #3?” (or any other specific number). This is a more specific question than the one actually asked. In the question, asked, there are six possible ways to fulfill the requirements of the problem, not one, because the problem does not specify whether the number should be 1, 2, 3, 4, 5, or 6.
Thus, any of the 6 red chips is acceptable for the first pick. However, on the second pick, only the blue chip with the same number as the red one that was just picked is acceptable (the chip must “match” the first one picked). Thus:
\(\frac{6}{12}\times \frac{1}{11}=\frac{1}{22}\).
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