1. **State the problem:** We are given that the probability a robot fails during any 6-hour shift is $p=0.10$. We want to find the probability that the robot operates through at most 5 shifts before it fails. This means the robot fails on or before the 5th shift. 2. **Identify the distribution:** This is a geometric distribution problem where the probability of failure (success in geometric terms) is $p=0.10$. The random variable $X$ represents the shift number on which the robot fails. 3. **Calculate the probability:** The probability that the robot fails on or before the 5th shift is $P(X \leq 5) = 1 - P(X > 5)$. 4. **Calculate $P(X > 5)$:** This is the probability the robot does not fail in the first 5 shifts, i.e., it operates successfully through 5 shifts: $$P(X > 5) = (1-p)^5 = (1-0.10)^5 = 0.9^5$$ 5. **Evaluate $0.9^5$:** $$0.9^5 = 0.59049$$ 6. **Find $P(X \leq 5)$:** $$P(X \leq 5) = 1 - 0.59049 = 0.40951$$ 7. **Compare with given options:** The closest option to $0.40951$ is none exactly, but since the problem asks for at most 5 shifts, the probability is approximately $0.4095$, which is not listed. However, if the problem intended "at most 5 shifts before it fails" to mean the robot fails within 5 shifts (including the 5th), then the answer is $0.4095$. If the problem meant the robot operates through at most 5 shifts before failing (i.e., fails on or before the 6th shift), then we calculate $P(X \leq 6) = 1 - 0.9^6 = 1 - 0.531441 = 0.468559$, which matches option b. **Final answer:** The probability that the robot will operate through at most 5 shifts before it fails is approximately $0.468$, which corresponds to option b.