Final answer:
The probability that Sal wins exactly one game out of two, with each win having a probability of 0.7, is 0.42. This is determined by adding the two scenarios where Sal wins the first game but loses the second, and where she loses the first game but wins the second.
Step-by-step explanation:
The question asks for the probability that Sal wins exactly one game out of two when the probability of winning a single game is 0.7. This is a binomial probability problem because there are only two outcomes (win or lose), the probability of success (winning) is constant (0.7), and each game is independent of the other.
To calculate the probability of winning exactly one game, we use the binomial probability formula: P(X = k) = C(n, k) * p^k * (1-p)^(n-k), where C(n, k) is the number of combinations of n items taken k at a time, p is the probability of success on a single trial, and X is the random variable representing the number of successes (wins in this case).
For Sal winning exactly one out of two games, the calculation is as follows:
- First Game Win, Second Game Lose: The probability is 0.7 * (1 - 0.7) = 0.21.
- First Game Lose, Second Game Win: The probability is (1 - 0.7) * 0.7 = 0.21.
Since these two events are mutually exclusive (they cannot happen at the same time), we add the probabilities together: 0.21 + 0.21 = 0.42. Therefore, the probability that Sal wins exactly one game is 0.42.