We are given the following probabilities:

• $P(A) = 0.2$
• $P(B) = 0.78$
• Events $A$ and $B$ are independent.

Now let's solve each part:

a) $P(A \cap B)$ (The probability of both events A and B occurring)

Since $A$ and $B$ are independent, the probability of both events occurring is the product of their individual probabilities: $P(A \cap B) = P(A) \times P(B)$ $P(A \cap B) = 0.2 \times 0.78 = 0.156$

b) $P(A \cup B)$ (The probability of either A or B occurring)

The formula for the union of two events is: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ Substitute the known values: $P(A \cup B) = 0.2 + 0.78 - 0.156 = 0.824$

c) $P(A \mid B)$ (The conditional probability of A given B)

The conditional probability is calculated as: $P(A \mid B) = \frac{P(A \cap B)}{P(B)}$ Substitute the known values: $P(A \mid B) = \frac{0.156}{0.78} = 0.2$

d) $P(A' \cup B)$ (The probability of either A not occurring or B occurring)

Using the complement rule, $P(A') = 1 - P(A)$: $P(A') = 1 - 0.2 = 0.8$ Now calculate $P(A' \cup B)$ using the formula for the union: $P(A' \cup B) = P(A') + P(B) - P(A' \cap B)$ Since $A'$ and $B$ are independent, $P(A' \cap B) = P(A') \times P(B)$: $P(A' \cap B) = 0.8 \times 0.78 = 0.624$ Now substitute into the formula: $P(A' \cup B) = 0.8 + 0.78 - 0.624 = 0.956$

Final answers:

a) $P(A \cap B) = 0.156$
b) $P(A \cup B) = 0.824$
c) $P(A \mid B) = 0.2$
d) $P(A' \cup B) = 0.956$

Would you like further details or clarification?

Here are 5 related questions for further practice:

1. What is the probability of neither A nor B occurring?
2. How would you find $P(B \mid A)$?
3. What is $P(A' \cap B')$?
4. If $P(A)$ were different, how would that affect $P(A \cup B)$?
5. What would change if events A and B were dependent?

Tip: For independent events, the intersection probability is simply the product of individual probabilities, making the calculations straightforward!