Understanding the Fair Coin Toss Experiment

When a fair coin is tossed three times, each toss has two equally likely outcomes: Heads (H) or Tails (T). The total number of possible outcomes is $2^3 = 8$. We can list all possible outcomes (the sample space, S):

• HHH
• HHT
• HTH
• THH
• HTT
• THT
• TTH
• TTT

Each outcome has a probability of $\frac{1}{8}$ since the coin is fair.

Determining Event A: Exactly Two Heads

Event A is defined as getting exactly two heads in three tosses. Let's identify the outcomes from the sample space that satisfy this condition:

• HHT (Two Heads, One Tail)
• HTH (Two Heads, One Tail)
• THH (Two Heads, One Tail)

There are 3 outcomes in Event A. Therefore, the probability of Event A, denoted as $P(A)$, is:

$P(A) = \frac{\text{Number of outcomes in A}}{\text{Total number of outcomes}} = \frac{3}{8}$

Determining Event B: At Most Two Tails

Event B is defined as getting at most two tails. This means getting zero tails, one tail, or two tails. Let's identify the outcomes:

• Zero tails: HHH
• One tail: HHT, HTH, THH
• Two tails: HTT, THT, TTH

The outcomes in Event B are {HHH, HHT, HTH, THH, HTT, THT, TTH}. There are 7 outcomes in Event B.

Alternatively, we can consider the complement of Event B. The complement of "at most two tails" is "exactly three tails". The only outcome with exactly three tails is TTT.

The probability of getting exactly three tails is $\frac{1}{8}$ (since TTT is one outcome out of 8).

So, the probability of Event B, $P(B)$, is:

$P(B) = 1 - P(\text{exactly 3 tails}) = 1 - \frac{1}{8} = \frac{7}{8}$

Calculating the Union Probability P(A $\cup$ B)

We need to find the probability of the union of events A and B, denoted as $P(A \cup B)$. The formula for the union of two events is:

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$

First, let's find the intersection of A and B ($A \cap B$). This represents the outcomes that are common to both Event A (exactly two heads) and Event B (at most two tails).

• Outcomes in A: {HHT, HTH, THH}
• Outcomes in B: {HHH, HHT, HTH, THH, HTT, THT, TTH}

The common outcomes are {HHT, HTH, THH}. These are the outcomes with exactly two heads, and they also satisfy the condition of having at most two tails (specifically, one tail each).

So, $A \cap B$ = {HHT, HTH, THH}. The number of outcomes in $A \cap B$ is 3.

The probability of the intersection, $P(A \cap B)$, is:

$P(A \cap B) = \frac{3}{8}$

Now, we can substitute the probabilities into the union formula:

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$

$P(A \cup B) = \frac{3}{8} + \frac{7}{8} - \frac{3}{8}$

$P(A \cup B) = \frac{3 + 7 - 3}{8} = \frac{7}{8}$

Alternatively, we can list all the unique outcomes that belong to either A or B (or both):

• Outcomes in A: {HHT, HTH, THH}
• Outcomes in B: {HHH, HHT, HTH, THH, HTT, THT, TTH}
• The union $A \cup B$ contains all outcomes from A and B, without repetition: {HHH, HHT, HTH, THH, HTT, THT, TTH}

There are 7 outcomes in the union $A \cup B$. Thus, the probability $P(A \cup B)$ is:

$P(A \cup B) = \frac{7}{8}$