Events $$ A $$ and $$ B $$ are mutually exclusive. Suppose event $$ A $$ occurs with probability. 0.12 and event $$ B $$ occurs with probability 0.42 . Compute the following. (If necessary, consult a list of formulas.) (a) Compute the probability that $$ B $$ occurs or $$ A $$ does not occur (or both). (b) Compute the probability that either $$ A $$ occurs without $$ B $$ occurring or $$ B $$ occurs without $$ A $$ occurring. $$ \square $$

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Let's analyze the given problem step by step.

**Given:**
- Events $$ A $$ and $$ B $$ are mutually exclusive, meaning $$ A \cap B = \emptyset $$.
- $$ P(A) = 0.12 $$
- $$ P(B) = 0.42 $$

### Part (a)
**Compute the probability that $$ B $$ occurs or $$ A $$ does not occur (or both).**

Since $$ A $$ and $$ B $$ are mutually exclusive:
- $$ P(A \cup B) = P(A) + P(B) = 0.12 + 0.42 = 0.54 $$

Now, the event " $$ B $$ occurs or $$ A $$ does not occur" can be written as $$ B \cup A^c $$.

Given that $$ A $$ and $$ B $$ are mutually exclusive, $$ B \subseteq A^c $$. Therefore:
- $$ B \cup A^c = A^c $$

So:
$$
P(B \cup A^c) = P(A^c) = 1 - P(A) = 1 - 0.12 = 0.88
$$

**Answer for Part (a):** $$ 0.88 $$

### Part (b)
**Compute the probability that either $$ A $$ occurs without $$ B $$ occurring or $$ B $$ occurs without $$ A $$ occurring.**

The event can be expressed as $$ (A \cap B^c) \cup (B \cap A^c) $$.

Since $$ A $$ and $$ B $$ are mutually exclusive:
- $$ A \cap B^c = A $$
- $$ B \cap A^c = B $$

Thus:
$$
P\left((A \cap B^c) \cup (B \cap A^c)\right) = P(A) + P(B) = 0.12 + 0.42 = 0.54
$$

**Answer for Part (b):** $$ 0.54 $$

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**Final Answers:**

(a) The probability is 0.88;
(b) The probability is 0.54.
(a) The probability is 0.88. (b) The probability is 0.54.

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Part (b)

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