1. The problem asks for the probability that in a sample of 25 tablets, two or more are defective, given a 5% defect rate per tablet. 2. Let $X$ be the number of defective tablets in the sample. $X$ follows a binomial distribution with parameters $n=25$ and $p=0.05$. 3. We want $P(X \geq 2)$. This is easier to find using the complement rule: $$P(X \geq 2) = 1 - P(X < 2) = 1 - P(X=0) - P(X=1)$$ 4. Calculate $P(X=0)$: $$P(X=0) = \binom{25}{0} (0.05)^0 (0.95)^{25} = 1 \times 1 \times 0.95^{25} = 0.95^{25}$$ 5. Calculate $P(X=1)$: $$P(X=1) = \binom{25}{1} (0.05)^1 (0.95)^{24} = 25 \times 0.05 \times 0.95^{24}$$ 6. Compute the values: $$0.95^{25} \approx 0.2776$$ $$0.95^{24} \approx 0.2922$$ $$P(X=1) = 25 \times 0.05 \times 0.2922 = 1.25 \times 0.2922 = 0.3653$$ 7. Now find $P(X \geq 2)$: $$P(X \geq 2) = 1 - 0.2776 - 0.3653 = 1 - 0.6429 = 0.3571$$ 8. Therefore, the probability that two or more tablets are defective in the sample is approximately $0.3571$. Final answer: **0.3571**