Let the random variable \(X\) represent the number of pears. Each box has a probability of \(\frac{4}{11}\) of containing a pear.

The expected number of pears, \(\mu\), is given by:

\(\mu = 121 \times \frac{4}{11} = 44\)

The variance, \(\sigma^2\), is given by:

\(\sigma^2 = 121 \times \frac{4}{11} \times \frac{7}{11} = 28\)

We use a normal approximation to the binomial distribution. The standard deviation \(\sigma\) is:

\(\sigma = \sqrt{28}\)

We apply continuity correction for \(X < 39\), using \(X < 38.5\).

Standardizing, we find:

\(Z = \frac{38.5 - 44}{\sqrt{28}} = -1.039\)

Using the standard normal distribution table, we find:

\(P(Z < -1.039) = \Phi(-1.039) = 1 - 0.8506 = 0.149\)