the 69% of “ none” students who fail, i.e., P(F | none) = 0.69. We need to calculate this by accounting for the students who fail in the three different categories (events) given in the table: P(F | none) = 0.69 (since from the table, 31% of those those students who do not use the resource room pass)īut for the denominator P(F) we need to overall percentage of students who fail, which is not immediately given in the table. We can calculate the numerator as follows: We can do this using Bayes’ Theorem! Recall that Bayes’ Theorem gives us a way of calculating conditional probability: Bayes TheoremĪpplying this to P(not R | fail) gives us: Or, if we use “ none” in place of “ not R” (to match the label in the table), and use “ F” for “ fail“, we need to calculate: We can rephrase this as “what is the probability that a randomly selected student did not study in the resource room given that the student did not pass the course?”, i.e., we need to calculate the conditional probability Now let’s look at what the exercise is asking us for: “If a randomly chosen student did not pass the course, what is the probability that he or she did not study in the resource room?” (This should make sense from reading the first two sentences! If 66% of students spend time in the resource room, and half of those spend more than 90 minutes, then it should be clear that 33% of all students spend more than 90 minutes in the resource room.) Then by the Multiplication Rule, we can compute P( > 90 ): “ > 90” = “speends more than 90mins per week in the resource room” The first step, as usual, is to write down the given probabilities in terms of the following events: Here is a snapshot of Exercise #8 on HW7-ConditionalProbability: HW7 #8
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