1. The problem asks to find the probabilities (A) $P(M \cap S)$ and (B) $P(R)$ using the given probability tree. 2. From the tree, the first split is between $M$ and $R$ with probabilities $0.7$ and $0.3$ respectively. 3. From $M$, the branches split into $S$ and $R$ with probabilities $0.7$ and $0.3$ respectively. 4. From $R$, the branches split into $S$ and $R$ with probabilities $0.5$ and $0.5$ respectively. 5. To find $P(M \cap S)$, multiply the probability of $M$ by the probability of $S$ given $M$: $$P(M \cap S) = P(M) \times P(S|M) = 0.7 \times 0.7 = 0.49$$ 6. To find $P(R)$, consider both ways to get $R$: - Directly from the first split: $P(R) = 0.3$ - From $M$ branch to $R$: $P(M) \times P(R|M) = 0.7 \times 0.3 = 0.21$ - From $R$ branch to $R$: $P(R) \times P(R|R) = 0.3 \times 0.5 = 0.15$ 7. Total $P(R)$ is the sum of all paths ending in $R$: $$P(R) = 0.3 + 0.21 + 0.15 = 0.66$$ Final answers: (A) $P(M \cap S) = 0.49$ (B) $P(R) = 0.66$