The image shows a probability tree diagram and questions based on conditional probabilities. The tree diagram depicts an experiment with two trials, and there are branches for different outcomes (E and $E^c$) starting from nodes A, B, and C.

Let's solve the given questions step-by-step.

(a) $P(A)$

From the tree diagram, $P(A)$ is the probability of the first branch leading to A. The corresponding probability is 0.3.

So, $P(A) = 0.3$.

(b) $P(E \mid A)$

This is the conditional probability of $E$ given $A$. From the branch starting at A, the probability of $E$ is 0.5.

So, $P(E \mid A) = 0.5$.

(c) $P(A \cap E)$

This is the joint probability of both A and E happening. You calculate this by multiplying the probabilities along the path for A and E: $P(A \cap E) = P(A) \times P(E \mid A) = 0.3 \times 0.5 = 0.15$

(d) $P(E)$

To find the total probability of $E$, we need to sum the joint probabilities for all paths that end in $E$. From the diagram, the paths for $E$ are:

• A and E: $P(A \cap E) = 0.15$
• B and E: $P(B) \times P(E \mid B) = 0.4 \times 0.3 = 0.12$
• C and E: $P(C) \times P(E \mid C) = 0.3 \times 0.7 = 0.21$

So, $P(E) = 0.15 + 0.12 + 0.21 = 0.48$.

(e) Does $P(A \cap E) = P(A) \times P(E)$?

We already know that $P(A \cap E) = 0.15$, and now we calculate $P(A) \times P(E)$: $P(A) \times P(E) = 0.3 \times 0.48 = 0.144$

Since $0.15 \neq 0.144$, the answer is No.

(f) Are A and E independent events?

Two events are independent if $P(A \cap E) = P(A) \times P(E)$. From part (e), we found that $P(A \cap E) \neq P(A) \times P(E)$, so A and E are not independent.

The answer is No.

Would you like further details on any part of this solution?

Here are 5 related questions for deeper understanding:

1. What is conditional probability, and how is it calculated?
2. How do we find the probability of multiple independent events?
3. Can you explain the concept of joint probability in more detail?
4. What are mutually exclusive events, and how do they differ from independent events?
5. How would you extend this tree diagram to three trials instead of two?

Tip: Always verify if events are independent by comparing the joint probability with the product of individual probabilities.