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A survey of mathematics students at the college revealed that 44% consistently spent at least 1.5 hours on mathematics homework and 56% spent less. Of those who spent at least 1.5 hours on homework, 76% made an A or B in the course. Of those who spent less than 1.5 hours, 27% made an A or B. A student made an A or B in the course. Find the probability that the student spent at least 1.5 hours on homework.

I know that this is some type of Bayes rule problem. How can I apply the numbers to the rule?

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  • $\begingroup$ What is the probability of a student chosen at random spent at least 1.5 hours on homework AND got an A or B? What is the probability that one student chosen at random spent less than 1.5 hours on homework AND got an A or B? (conditional probabilies) $\endgroup$
    – Kevin
    Nov 28 '13 at 0:42
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Let $X$ be the event that a given student spends at least $1.5$ hours on homework.

Let $Y$ be the event that a given student earns an A or a B in the course.

We are given the following:

$$P(X) = 44\%$$ $$P(\neg X) = 56\%$$ $$P(Y\text{ }|\text{ }X) = 76\%$$ $$P(Y\text{ }|\text{ }\neg X) = 27\%$$

We are asked for the probability that a student spent at least $1.5$ hours on homework, given that he earned an A or a B in the course. We want to find $P(X\text{ }|\text{ }Y)$.

First we find $P(Y)$. We know that $$P(Y) = P(Y\text{ }|\text{ }X) \cdot P(X) + (Y\text{ }|\text{ }\neg X) \cdot P(\neg X)$$

We plug in to solve for $P(Y)$:

$$P(Y) = (0.76)(0.44) + (0.27)(0.56) = 0.4856$$

Bayes' Theorem states that $$P(X\text{ }|\text{ }Y) = \frac{P(Y\text{ }|\text{ }X) \cdot P(X)}{P(Y)}$$

Now we can just plug in the known values to get $$P(X\text{ }|\text{ }Y) = \frac{(0.76)(0.44)}{0.4856} = \frac{418}{607} \approx 68.86\%$$

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