Final answer:
The probability of losing the first three games and winning the fourth game, assuming each game has an equal chance of winning or losing (0.5), is calculated by multiplying the probabilities of each independent event, resulting in A)1/16.
Step-by-step explanation:
The question asks about the probability of a sequence of outcomes for playing a series of games. Specifically, the scenario is losing the first three games and winning the fourth game. Assuming that each game is independent and has only two outcomes (win or lose), the probability of each outcome is not given but typically it is assumed that the chance of winning is equal to the chance of losing, hence 0.5 for each. Therefore, we can calculate the probability of this sequence by multiplying the probabilities of each independent event.
Probability of losing the first game (L) = 0.5
Probability of losing the second game (L) = 0.5
Probability of losing the third game (L) = 0.5
Probability of winning the fourth game (W) = 0.5
The combined probability of the sequence LLLW is the product of each individual probability: P(LLLW) = P(L) × P(L) × P(L) × P(W) = 0.5 × 0.5 × 0.5 × 0.5 = 0.5^4 = 0.0625. Thus, the correct answer is 1/16, which corresponds to option A).