Question

Davis plays two games.

What is the probability that he wins exactly one of the games?
Give your answer as a fraction in its simplest form.
Game 1
8
11
3
11
Win
Lose
Game 2
8
11
8
3
11
8
11
3
11
Win
Lose
Win
Lose

Answer 1

The probability that Davis wins exactly one of the two games is the sum of the probabilities of winning one game and losing the other, resulting in a probability of 48/121.

Step-by-step explanation:

The student is asking about the probability of winning exactly one game when two games are played with known probabilities of winning or losing each. To find the probability of winning exactly one game, you need to consider the two possible scenarios:

• Davis wins Game 1 and loses Game 2.
• Davis loses Game 1 and wins Game 2.

For the first scenario, the probability is the product of the probability of winning Game 1 and the probability of losing Game 2: (8/11) × (3/11).

For the second scenario, the probability is the product of the probability of losing Game 1 and the probability of winning Game 2: (3/11) × (8/11).

Since these two events are mutually exclusive, you can simply add the probabilities of the two scenarios together:

Probability of winning exactly one game = (8/11) × (3/11) + (3/11) × (8/11) = 24/121 + 24/121 = 48/121.

This fraction is already in its simplest form.

Answer 2

Step-by-step explanation:

To find the probability that he wins exactly one of the games, we need to count the number of outcomes in which he wins one game and loses the other game, and then divide that by the total number of possible outcomes.

The outcomes in which he wins one game and loses the other game are:

Win Game 1, Lose Game 2

Lose Game 1, Win Game 2

The probability of winning Game 1 is 2/4, or 1/2. The probability of losing Game 1 is therefore 1 - 1/2 = 1/2. Similarly, the probability of winning Game 2 is 3/6, or 1/2, and the probability of losing Game 2 is therefore 1 - 1/2 = 1/2.

So, the probability of winning one game and losing the other is:

(1/2) * (1/2) + (1/2) * (1/2) = 1/2

Therefore, the probability that he wins exactly one of the games is 1/2.