Let the events of Liam winning the two games be denoted as $A$ (for game 1) and $B$ (for game 2). The key pieces of information are:

• The probability that Liam wins at least one game is 80%, or $P(A \cup B) = 0.8$.
• Liam is equally likely to win each game, meaning $P(A) = P(B) = p$.

Step 1: Use the formula for the union of two events.

The probability of the union of two events can be written as: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ Substitute the known values: $0.8 = p + p - P(A \cap B)$ which simplifies to: $0.8 = 2p - P(A \cap B)$

Step 2: Probability of winning both games.

The probability of winning both games is $P(A \cap B)$, and this is what we need to find. Rearranging the equation: $P(A \cap B) = 2p - 0.8$

Step 3: Use additional information.

Since Liam is equally likely to win each game, $P(A) = P(B) = p$. The probability of winning at least one game is 80%, which means the probability of losing both games is $1 - 0.8 = 0.2$.

The probability of losing both games is $(1 - p)(1 - p) = (1 - p)^2$. Therefore, we have: $(1 - p)^2 = 0.2$

Step 4: Solve for $p$.

Take the square root of both sides: $1 - p = \sqrt{0.2} \approx 0.447$ Thus: $p = 1 - 0.447 = 0.553$

Step 5: Calculate the probability of winning both games.

Substitute $p = 0.553$ into the equation for $P(A \cap B)$: $P(A \cap B) = 2(0.553) - 0.8 = 1.106 - 0.8 = 0.306$

Thus, the probability that Liam will win both games is approximately $0.306$, or 30.6%.

Do you want any more details or have questions?

Here are 5 related questions you might find interesting:

1. What is the probability that Liam loses both games?
2. What is the probability that Liam wins exactly one game?
3. How would the result change if the probability of winning each game were different?
4. Can this be represented using a Venn diagram for better visualization?
5. How can the result be verified using a simulation?

Tip: Remember that probabilities must sum to 1. This principle is useful when checking your calculations.