ExplanationThe probability of any event equals the number of ways to get the desired outcome divided by the total number of outcomes.

Start with the denominator, which is the total number of ways that the principal can choose two children from the classroom. Use the fundamental counting principle. There are 6 possible options for the first choice and 5 for the second, giving (6)(5) = 30 possibilities.

However, this double-counts some cases; for example, choosing Jan and then Robert is the same as choosing Robert and then Jan.

Divide the total number of pairs by 2: \(\frac{6 \times 5}{2}= 1\). Alternatively, use the formula for a set in which the order doesn’t matter: \(\frac{total!}{in! \times out!}\).

In this case: \(\frac{6!}{2! \times 4!}=15\).

Now compute the numerator, which is the number of pairs that include Jan. Since the pair only includes two children and one is already decided (Jan), there are exactly 5 options for the other child. Thus, there are 5 total pairs that include Jan: Jan with each of the other students.

The probability of choosing a pair with Jan is \(\frac{1}{3}\).

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Sandy

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