Archie is taking a course in probability theory. Discussion sessions with the class TA (teaching assistant) conflict with Archie's trips to the beach. Archie has to decide between attending TA sessions, which might help him do better in the probability course, and going to the beach.
Archie learns that 93% of past students who did not attend TA sessions received a grade of B or below in the course. Archie also learns that 11% of past students received a grade higher than B. After some thought, Archie decides to attend the TA sessions. Not all of his classmates share his decision; in fact, only 4 out of every 18 of his classmates attend discussion sessions with the TA. Compute the probability that Archie will receive a grade higher than a B in the course.

We are given that Archie learns that 93% of past students who did not attend TA sessions received a grade of B or below in the course. Archie also learns that 11% of past students received a grade higher than B.

Let B = event that the student received a grade higher than B.

and T = event that the student discussion sessions with the TA.

So, P(B) = 0.11 and P(T) = [tex]\frac{4}{18}[/tex] = 0.22

Also, the probability that the students who did not attend TA sessions received a grade of B or below in the course = P(B'/T') = P(not B/not T) = 0.92

Now, we have to find the probability that Archie will receive a grade higher than a B in the course given he attended the discussion sessions with the TA = P(B/T)

As we know that the conditional probability formula is given by;

P(B/T) = [tex]\frac{\text{P}(\text{B} \bigcap \text{T})}{\text{P(T)}}[/tex]

Now, as it is given that P(not B/not T) = 0.92

So, [tex]\frac{\text{P}(\text{not B} \bigcap \text{not T})}{\text{P(not T)}} =0.92[/tex]

[tex]\frac{\text{P}(\text{not B} \bigcap \text{not T})}{1 - \text{P(T)}} =0.92[/tex]

[tex]\frac{\text{P}(\text{not B} \bigcap \text{not T})}{1 - 0.22}} =0.92[/tex]

[tex]{\text{P}(\text{not B} \bigcap \text{not T})} =0.92\times 0.78[/tex]

[tex]{\text{P}(\text{not B} \bigcap \text{not T})}[/tex] = 0.7176

Also, [tex]{\text{P}(\text{ B} \bigcup \text{ T})}= 1 - {\text{P}(\text{not B} \bigcap \text{not T})}[/tex]

[tex]{\text{P}(\text{ B} \bigcup \text{ T})}[/tex] = 1 - 0.7176 = 0.2824

Now, the formula for [tex]{\text{P}(\text{ B} \bigcup \text{ T})}[/tex] is given by;

[tex]{\text{P}(\text{ B} \bigcup \text{ T})} = \text{P(B)} + \text{P(T)} - {\text{P}(\text{ B} \bigcap \text{ T})}[/tex]

0.2824 = 0.11 + 0.22 - [tex]\text{P(B} \bigcap \text{T})[/tex]

[tex]\text{P(B} \bigcap \text{T})[/tex] = 0.33 - 0.2824 = 0.0476

Now, the required probability P(B/T) = [tex]\frac{\text{P}(\text{B} \bigcap \text{T})}{\text{P(T)}}[/tex]

= [tex]\frac{0.0476}{0.22}[/tex] = 0.2164

Hence, the probability that Archie will receive a grade higher than a B in the course is 0.2164.