Here’s the breakdown of the problem based on the provided image:

We are tasked to find the following probabilities:

1. $P(B' \cap A)$
2. $P(A \cap B)$
3. $P(A' \cap B')$

Given Information:

• $P(A) = 0.35$
• $P(B) = 0.5$
• $P(A' \cap B) = 0.3$

Solution Steps:

1. $P(B' \cap A)$:

Using the complement rule, we know: $P(B' \cap A) = P(A) - P(A \cap B).$

To solve this, we need $P(A \cap B)$.

2. $P(A \cap B)$:

From the given data, we can calculate: $P(A \cap B) = P(A) - P(A' \cap B),$ since $P(A) = P(A \cap B) + P(A' \cap B)$.

Substitute: $P(A \cap B) = 0.35 - 0.3 = 0.05.$

3. $P(B' \cap A)$ (continued):

Now substitute $P(A \cap B)$ into the equation: $P(B' \cap A) = P(A) - P(A \cap B) = 0.35 - 0.05 = 0.3.$

4. $P(A' \cap B')$:

Using the inclusion-exclusion principle, we know: $P(A' \cap B') = 1 - P(A \cup B).$

Also, $P(A \cup B)$ can be written as: $P(A \cup B) = P(A) + P(B) - P(A \cap B).$

Substitute values: $P(A \cup B) = 0.35 + 0.5 - 0.05 = 0.8.$

Thus: $P(A' \cap B') = 1 - 0.8 = 0.2.$

Final Results:

1. $P(B' \cap A) = 0.3$
2. $P(A \cap B) = 0.05$
3. $P(A' \cap B') = 0.2$

Would you like a detailed explanation of any step?

Here are 5 related questions to further your understanding:

1. What is the complement rule, and how is it applied in probability problems?
2. Can you derive $P(A' \cap B')$ using other methods?
3. How does the inclusion-exclusion principle work in probability?
4. What other relationships exist between $P(A \cap B)$ and $P(A \cup B)$?
5. How can these concepts be extended to more than two events?

Tip: Drawing a Venn diagram can help visualize intersections and complements in probability problems!