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Thirty-five students in algebra 2 class took a test. 9 received A's, 18 received B's, and 8 received C's. If the teacher randomly chooses 3 tests, what is the probability that the teacher chose tests with grades A, B, and C in that order?

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Basic approach. There are $35$ tests in all. There are $35$ ways to select the first test, $x$ ways to select the second test (once the first selection has already removed that test from the pile), and $y$ ways to select the third test. Determine what $x$ and $y$ are. Then the total number of ways to select three tests, maintaining sequence, is $35 \times x \times y$.

Of those ways, only some satisfy the given condition. How many ways are there to do that? There are $9$ ways to select the first test appropriately (because there are $9$ tests that received an A), $z$ ways to select the second test appropriately (from the tests that received a B), and $w$ ways to select the third test appropriately. Thus, the total number of ways to select three tests that satisfy the condition, is $9 \times z \times w$.

The probability you want is then the number of satisfactory selections, divided by the number of total selections (satisfactory or otherwise), or

$$ \frac{35 \times x \times y}{9 \times z \times w} $$

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  • $\begingroup$ This completely makes sense, but wouldn't there be greater than a 3.3% chance of selecting an A, then a B, and then a C? Just seems low $\endgroup$ – L doe May 14 '18 at 2:20
  • $\begingroup$ @Ldoe: It's essentially in the neighborhood of a one in four chance, followed by a one in two chance, followed by another one in four chance. Individually, they don't seem that unlikely, but they combine to about a one in $32$ chance. $\endgroup$ – Brian Tung May 14 '18 at 18:13

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