To find the probability of losing the first three games and winning the fourth game, we need to know the probability of losing a single game and the probability of winning a single game.
Let's assume that the probability of losing a single game is \(p\) and the probability of winning a single game is \(q\). Since winning and losing are mutually exclusive events, \(p + q = 1\).
To calculate the probability of losing the first three games and winning the fourth game, we multiply the probabilities of each individual event together:
Probability = Probability of losing the first game * Probability of losing the second game * Probability of losing the third game * Probability of winning the fourth game
Since the events are independent, we can multiply the probabilities:
Probability = \(p \times p \times p \times q = p^3 \times q\)
So, the probability of losing the first three games and winning the fourth game is \(p^3 \times q\).
Please note that without specific values for \(p\) and \(q\), we cannot provide an exact numerical answer.