Let the event that the pen is in the pencil case be denoted as \(P(C) = 0.7\), and the event that Ted finds the pen given it is in the pencil case be \(P(F|C) = 1\). The event that the pen is not in the pencil case is \(P(C') = 0.3\), and the probability that Ted finds the pen given it is not in the pencil case is \(P(F|C') = 0.2\).

We need to find \(P(C|F)\), the probability that the pen is in the pencil case given that Ted finds it. Using Bayes' theorem:

\(P(C|F) = \frac{P(C \cap F)}{P(F)}\)

Where:

\(P(C \cap F) = P(C) \cdot P(F|C) = 0.7 \times 1 = 0.7\)

\(P(F) = P(C \cap F) + P(C' \cap F) = 0.7 + (0.3 \times 0.2) = 0.7 + 0.06 = 0.76\)

Thus:

\(P(C|F) = \frac{0.7}{0.76} = 0.921\)