The diagram you provided seems to depict a probability tree with events $A$ and $B$ leading to outcomes $C$ and $D$. The probabilities on the branches represent conditional probabilities.

Here’s how to approach solving for the outcomes:

1. $P(A \cap C)$: This represents the probability of going through event $A$ and then to event $C$. To find this: $P(A \cap C) = P(A) \times P(C|A)$ From the diagram: $P(A) = 0.3, \quad P(C|A) = 0.4$ Therefore, $P(A \cap C) = 0.3 \times 0.4 = 0.12$

2. $P(A \cap D)$: This is the probability of going through event $A$ and then to event $D$. It follows the same rule: $P(A \cap D) = P(A) \times P(D|A)$ From the diagram: $P(A) = 0.3, \quad P(D|A) = 1 - 0.4 = 0.6$ So, $P(A \cap D) = 0.3 \times 0.6 = 0.18$

3. $P(B \cap C)$: This represents the probability of going through event $B$ and then to event $C$: $P(B \cap C) = P(B) \times P(C|B)$ From the diagram: $P(B) = 1 - 0.3 = 0.7, \quad P(C|B) = 0.1$ Therefore, $P(B \cap C) = 0.7 \times 0.1 = 0.07$

4. $P(B \cap D)$: This is the probability of going through event $B$ and then to event $D$: $P(B \cap D) = P(B) \times P(D|B)$ From the diagram: $P(B) = 0.7, \quad P(D|B) = 1 - 0.1 = 0.9$ So, $P(B \cap D) = 0.7 \times 0.9 = 0.63$

Final Probabilities:

• $P(A \cap C) = 0.12$
• $P(A \cap D) = 0.18$
• $P(B \cap C) = 0.07$
• $P(B \cap D) = 0.63$

Would you like any further details on these calculations?

Follow-up Questions:

1. How would you calculate $P(C)$ and $P(D)$ from this tree?
2. What would the total probability $P(C) + P(D)$ sum up to?
3. How could we extend this tree diagram if there were additional outcomes beyond $C$ and $D$?
4. Could this method be applied to real-life decision-making scenarios, such as medical testing or financial investments?
5. How would the probabilities change if event $A$ or $B$ were altered?

Tip:

Always ensure that the sum of probabilities at each branch adds up to 1.