Help Please. Conditional Probability Question You ask your neighbour to water 3 sickly plants in your front yard while you are on vacation. Each of the plants will be alive with probability 0.9 if watered; it will be alive with probability 0.3 if not watered. You are 95 percent certain that your neighbour will remember to water the plants. Let X be the number of the sickly plants that you will find to be alive when you return from the vacation. a) Find P(X=2) b) If only one plant is found to be alive when you return, what is the conditional probability that your neighbour remembered to water the plants? I got an answer of 0.2403 for (a) but using a tree diagram and would wonder how to do it using formulas

(a) $P(X = 2) = P(\text{two get watered})\cdot P(\text{two live when watered}) + P(\text{two don't get watered}) \cdot P(\text{two live without water})$

$$= (0.95) \cdot \left({3\choose 2}(0.9)^{2}(0.1)\right) + (0.05)\cdot \left({3\choose 2}(0.3)^{2}(0.7)\right) = \boxed{0.2403} \\\$$

$\\\$

(b) $P(\text{neighbor remembered } \mid \text{one plant alive}) = \frac{P(\text{one plant alive} \mid \text{neighbor remembered}) \cdot P(\text{neighbor remembered})}{P(\text{one plant alive})}$

$$= \frac{\left({3\choose 1} \cdot (0.9)^{1} \cdot (0.1)^{2}\right) \cdot \left(0.95\right)}{(0.95) \cdot \left({3\choose 1}(0.9)(0.1)^2\right) + (0.05)\cdot \left({3\choose 1}(0.3)^{1}(0.7)^2\right)} = \boxed{0.537735849}$$

Note: ${n\choose k} = \frac{n!}{k!(n - k)!}$