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Conditional probability explores the likelihood of an event occurring, given that another event has already occurred. In this driving test scenario, you're working with two tries: the first driving test and potentially a second test if the first attempt is unsuccessful. Understanding conditional probability is crucial because the event of passing on the second try is conditional on failing the first try. Here, we denote failing the test on the first attempt as the complement of the event of passing the first try, which is written as \( A^c \).The probability of passing on the second try, given the first try was failed, is denoted as \( P(B|A^c) \). - In this scenario, \( P(B|A^c) = 0.75 \), meaning there is a 75% chance to pass the test on the second try if you did not pass the first. This condition-based approach helps us refine our understanding of sequences of events, rather than thinking of each attempt in isolation. Using conditional probability simplifies the calculation by considering dependencies between events.

Complementary events encompass the idea that two outcomes are exhaustive alternatives of each other in any experiment. If one occurs, the other does not.In the exercise, we encounter the complementary event for passing the first test, termed as failing the test, represented by \( A^c \). - Since the probability of passing the test on the first try is \( P(A) = 0.60 \), it naturally follows that the probability of not passing on the first try is \( P(A^c) = 1 - P(A) = 0.40 \).Understanding complementary events is crucial because it allows us to determine the probability of passing the test on the second attempt. This concept ensures that all outcomes of an experiment are accounted for in probability calculations, and that no divergence or gaps exist in the analysis of such problems.

Probability calculations are essential for determining the likelihood of specific outcomes within a given scenario. In the driving test example, the calculation steps provide insight into how we sum probabilities to arrive at a comprehensive solution.The final goal is the total probability that a driver will pass without having to wait 6 months, expressed as \( P(A) + P(A^c) \times P(B|A^c) \).Each term in this formula is crucial:- \( P(A) = 0.60 \) is the probability of passing on the first try.- \( P(A^c) \) indicates the probability of failing the first try, which helps calculate chances for the second try.- \( P(B|A^c) = 0.75 \), representing the probability of passing the second attempt given the first was failed.By multiplying the complementary probability \( P(A^c) \) by \( P(B|A^c) \), and then adding it to \( P(A) \), we get \( 0.60 + 0.40 \times 0.75 = 0.90 \). This value represents the total probability of passing without the waiting period. This thorough calculation process empowers us to understand how distinct probabilities piece together, providing a clear path to understanding more complex probability scenarios efficiently.