in is tossed three times. Let A = {three heads occur} and B = {at least one head occurs}.
What is `" p (A cup B)"`?

Options

A. `1/8`
B. `1/4`
C. `1/2`
D. `7/8`

Explanation

To solve the problem, we need to find the probability of the union of two events A and B when a coin is tossed three times.

**Step 1: Identify the sample space.**
When a coin is tossed three times, the possible outcomes are:
1. HHH (3 heads)
2. HHT (2 heads, 1 tail)
3. HTH (2 heads, 1 tail)
4. THH (2 heads, 1 tail)
5. HTT (1 head, 2 tails)
6. THT (1 head, 2 tails)
7. TTH (1 head, 2 tails)
8. TTT (0 heads)

Thus, the sample space S consists of 8 outcomes: {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}.

**Step 2: Define events A and B.**
- Event A: {three heads occur} = {HHH}
- Event B: {at least one head occurs} = {HHH, HHT, HTH, THH, HTT, THT, TTH}

**Step 3: Calculate the probabilities of events A and B.**
- The probability of event A (P(A)):
- There is 1 favorable outcome (HHH) out of 8 possible outcomes.
- Thus, P(A) = 1/8.

- The probability of event B (P(B)):
- There are 7 favorable outcomes (HHH, HHT, HTH, THH, HTT, THT, TTH) out of 8 possible outcomes.
- Thus, P(B) = 7/8.

**Step 4: Find the intersection of events A and B.**
- The intersection of A and B (A ∩ B) consists of outcomes that satisfy both events:
- A = {HHH} and B = {HHH, HHT, HTH, THH, HTT, THT, TTH}
- The only outcome that satisfies both A and B is HHH.
- Thus, P(A ∩ B) = 1/8.

**Step 5: Use the formula for the probability of the union of two events.**
The formula for the probability of the union of two events is:
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]

Substituting the values we calculated:
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]
\[ P(A \cup B) = \frac{1}{8} + \frac{7}{8} - \frac{1}{8} \]
\[ P(A \cup B) = \frac{1 + 7 - 1}{8} = \frac{7}{8} \]

**Final Answer:**
The probability \( P(A \cup B) = \frac{7}{8} \).

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