Question

suppose your school is in the midst of a flu epidemic. the probability that a randomly-selected student has the flu is 0.30, and the probability that a student who has the flu also has a high fever is 0.90. but there are other illnesses making the rounds, and the probability that a student who doesn't have the flu does have a high fever (as a result of some other ailment) is 0.12. suppose a student walks into the nurse's office with a high fever. what is the probability that she has the flu?

Answer 1

Given a flu probability of 30%, high fever probability with flu of 90%, and high fever probability without flu of 12%, the probability that a student has the flu given a high fever is approximately 73.77%.

We have been given the following information:

`\( P(\text{Flu}) = 0.30 \)\\\( P(\text Flu) = 0.90 \)\\\( P(\text No Flu) = 0.12 \)`

Calculate
`\( P(\text{No Flu}) \)`:

`\[ P(\text{No Flu}) = 1 - P(\text{Flu}) = 1 - 0.30 = 0.70 \]`

Now, use the law of total probability to find
`\( P(\text{High Fever}) \)`:

`\[ P(\text{High Fever}) = P(\textHigh Fever ) \cdot P(\text{Flu}) + P(\textHigh Fever ) \cdot P(\text{No Flu}) \]`

`\[ P(\text{High Fever}) = 0.90 \cdot 0.30 + 0.12 \cdot 0.70 = 0.366 \]`

Now, use Bayes' Theorem to find
`\( P(\textFlu ) \)`:

`\[ P(\text High Fever) = \frac{P(\textHigh Fever ) \cdot P(\text{Flu})}{P(\text{High Fever})} \]\[ P(\textFlu ) = (0.90 \cdot 0.30)/(0.366) \approx 0.7377 \]`

Therefore, the probability that the student has the flu given that she has a high fever is approximately 0.7377, or 73.77%.