1. **Problem statement:** (a) Given $P(A) = 0.9$ and $P(L) = 0.45$, find the probability that a worker with a laptop is connected to router A. (b) Given $P(L) = 0.45$, $P(D) = 0.30$, and $P(L \cap D) = 0.18$, find the percentage of workers who have a desktop and also have a laptop. 2. **Formulas and rules:** - For (a), assuming independence, the probability that a worker has a laptop and is connected to router A is $P(A \cap L) = P(A) \times P(L)$. - For (b), the conditional probability formula is $P(L|D) = \frac{P(L \cap D)}{P(D)}$. 3. **Calculations:** (a) Calculate $P(A \cap L)$: $$P(A \cap L) = P(A) \times P(L) = 0.9 \times 0.45 = 0.405$$ So, the probability that a worker with a laptop is connected to router A is $0.405$ or 40.5%. (b) Calculate $P(L|D)$: $$P(L|D) = \frac{P(L \cap D)}{P(D)} = \frac{0.18}{0.30} = 0.6$$ Convert to percentage: $$0.6 \times 100 = 60\%$$ So, 60% of those who have a desktop also have a laptop. 4. **Summary:** - (a) Probability of a worker with a laptop connected to router A is 40.5%. - (b) 60% of desktop owners also have a laptop.