To solve the problem, I'll explain how to calculate the probabilities $P(M \cap S)$ and $P(R)$ using the information provided and a probability tree structure.

Part A: $P(M \cap S)$

Assuming that $M$ and $S$ are events in a probability tree, where the branches indicate conditional probabilities:

• Let's say the first branch splits into $M$ and $R$ with probabilities 0.8 and 0.2, respectively.
• Suppose after $M$, there are two further branches representing $S$ and $\text{not } S$ with probabilities 0.5 and 0.5.
• After $R$, the branches represent $S$ and $\text{not } S$ with probabilities 0.4 and 0.6.

Given this information, the probability of both $M$ and $S$ occurring is calculated by multiplying the probability along the path $M$ followed by $S$:

$P(M \cap S) = P(M) \times P(S \mid M)$

$P(M \cap S) = 0.8 \times 0.5 = 0.4$

So,

$P(M \cap S) = 0.4$

Part B: $P(R)$

The probability $P(R)$ is directly given as the probability of the branch corresponding to $R$:

$P(R) = 0.2$

Summary of Results:

• (A) $P(M \cap S) = 0.4$
• (B) $P(R) = 0.2$

Would you like more details, or do you have any questions?

Here are some related questions you might consider:

1. How would you calculate $P(\text{not } M \cap S)$?
2. What is the probability $P(\text{not } M \cup S)$?
3. If $P(M)$ were different, how would that affect $P(M \cap S)$?
4. Can you explain how the sum of $P(M \cap S)$ and $P(\text{not } M \cap S)$ relate to $P(S)$?
5. What is the probability of $S$ given that $R$ occurs?
6. How does the tree structure help in calculating conditional probabilities?
7. Can you describe the concept of independent events in this context?
8. What would change if the branches for $S$ and $\text{not } S$ were reversed?

Tip: When using a probability tree, ensure that the probabilities of all branches stemming from a single point add up to 1.