The probability of at most 5 defective fuses in a box of 200 fuses is 0.79. This can be calculated by using the binomial distribution. The binomial distribution is a probability distribution that describes the number of successes in a sequence of independent experiments each of which yields success with probability $p$. In this case, the experiment is selecting a fuse from the box, and the success is selecting a defective fuse. The probability of selecting a defective fuse is 0.02, since 2% of the fuses are defective.
The binomial distribution can be used to calculate the probability of any number of successes in a sequence of experiments. In this case, we are interested in the probability of selecting at most 5 defective fuses. This can be calculated by adding up the probabilities of selecting 0, 1, 2, 3, 4, and 5 defective fuses.
The probability of selecting 0 defective fuses is $0.98^{200} \approx 0.368$. The probability of selecting 1 defective fuse is $200 \cdot 0.02 \cdot 0.98^{199} \approx 0.304$. The probability of selecting 2 defective fuses is $200 \cdot 0.02^2 \cdot 0.98^{198} \approx 0.128$. The probability of selecting 3 defective fuses is $200 \cdot 0.02^3 \cdot 0.98^{197} \approx 0.040$. The probability of selecting 4 defective fuses is $200 \cdot 0.02^4 \cdot 0.98^{196} \approx 0.008$. The probability of selecting 5 defective fuses is $200 \cdot 0.02^5 \cdot 0.98^{195} \approx 0.001$.
Adding up these probabilities, we get $0.368 + 0.304 + 0.128 + 0.040 + 0.008 + 0.001 = 0.79$.
Therefore, the probability of at most 5 defective fuses in a box of 200 fuses is 0.79.