A gardener wants to plant saplings on a day when rain is not predicted. According to the forecast…

Let $R_1$ be the event that it rains today and $R_2$ be the event that it rains tomorrow. Given: $P(R_1) = 0\cdot 4$, $P(R_1') = 1 - 0\cdot 4 = 0\cdot 6$. $P(R_2 \mid R_1) = 0\cdot 8$, $P(R_2 \mid R_1') = 0\cdot 7$. The gardener does not plant saplings tomorrow if it rains tomorrow ($R_2$). By total probability: $P(R_2) = P(R_1)P(R_2 \mid R_1) + P(R_1')P(R_2 \mid R_1') = (0\cdot 4)(0\cdot 8) + (0\cdot 6)(0\cdot 7) = 0\cdot 32 + 0\cdot 42 = 0\cdot 74$.

The gardener will plant saplings tomorrow if it does not rain tomorrow ($R_2'$). $P(R_2') = 1 - P(R_2) = 1 - 0\cdot 74 = 0\cdot 26$.

We require the probability that he did not plant today ($R_1$, rain today) given that he does not plant tomorrow ($R_2$, rain tomorrow). By Bayes' theorem: $P(R_1 \mid R_2) = \frac{P(R_1)P(R_2 \mid R_1)}{P(R_2)} = \frac{0\cdot 32}{0\cdot 74} = \frac{32}{74} = \frac{16}{37} \approx 0\cdot 432$.

He can plant saplings on both days if it does not rain on either day ($R_1' \cap R_2'$). $P(R_1' \cap R_2') = P(R_1') P(R_2' \mid R_1') = (0\cdot 6)(1 - 0\cdot 7) = (0\cdot 6)(0\cdot 3) = 0\cdot 18$.