HELP PLEASE SO I CAN UNDERSTAND

Question

Two teaching methods for a statistics class, online and face-to-face, are available during the course of an academic year. To date, 11 out of every 25 students has failed the online class and 33 out of every 50 students have failed the face-to-face class. However, the online class is only offered 21%21% of the time due to faculty constraints and the face-to-face class is offered the other 79%79% of the time. What is the probability that a student who failed the course took the class online? Round your answer to four decimal places. (Do not round intermediate values.)

Answer 1

To find the probability that a student who failed the course took the class online, we can use Bayes' theorem. Let's define the relevant variables:

1. Let OOO be the event that a student took the online class.
2. Let FFF be the event that a student failed the course.

We want to calculate P(O∣F)P(O | F)P(O∣F), the probability that a student took the online class given that they failed.

Step 1: Gather the information

1. Failure Rates:
2. For the online class:
3. P(F∣O)=1125=0.44P(F | O) = \frac{11}{25} = 0.44P(F∣O)=2511​=0.44
4. For the face-to-face class:
5. P(F∣Face-to-Face)=3350=0.66P(F | \text{Face-to-Face}) = \frac{33}{50} = 0.66P(F∣Face-to-Face)=5033​=0.66
6. Proportions of Classes Offered:
7. Online: P(O)=0.21P(O) = 0.21P(O)=0.21
8. Face-to-Face: P(Face-to-Face)=0.79P(\text{Face-to-Face}) = 0.79P(Face-to-Face)=0.79

Step 2: Calculate the total probability of failing the course, P(F)P(F)P(F)

Using the law of total probability:

P(F)=P(F∣O)⋅P(O)+P(F∣Face-to-Face)⋅P(Face-to-Face)P(F) = P(F | O) \cdot P(O) + P(F | \text{Face-to-Face}) \cdot P(\text{Face-to-Face})P(F)=P(F∣O)⋅P(O)+P(F∣Face-to-Face)⋅P(Face-to-Face)

Substituting the values:

P(F)=(0.44⋅0.21)+(0.66⋅0.79)P(F) = (0.44 \cdot 0.21) + (0.66 \cdot 0.79)P(F)=(0.44⋅0.21)+(0.66⋅0.79)

Step 3: Calculate P(O∣F)P(O | F)P(O∣F)

Using Bayes' theorem:

P(O∣F)=P(F∣O)⋅P(O)P(F)P(O | F) = \frac{P(F | O) \cdot P(O)}{P(F)}P(O∣F)=P(F)P(F∣O)⋅P(O)​