Let the probability that Khalid rides his skateboard on a given day be 0.4. We need to find the probability that he rides on his skateboard on at least 2 out of 10 days.

This is a binomial probability problem where the number of trials is 10, and the probability of success (riding the skateboard) is 0.4.

The probability of riding on the skateboard on 0 or 1 day is:

\(P(0) = \binom{10}{0} (0.4)^0 (0.6)^{10} = (0.6)^{10}\)

\(P(1) = \binom{10}{1} (0.4)^1 (0.6)^9 = 10 \times 0.4 \times (0.6)^9\)

Thus, the probability of riding on the skateboard on at least 2 days is:

\(1 - (P(0) + P(1)) = 1 - [(0.6)^{10} + 10 \times 0.4 \times (0.6)^9]\)

Calculating these values:

\((0.6)^{10} \approx 0.0060466176\)

\(10 \times 0.4 \times (0.6)^9 \approx 0.0403936\)

So:

\(1 - (0.0060466176 + 0.0403936) = 1 - 0.04644 \approx 0.954\)