We are tasked with finding the probability that Chang will receive a grade higher than a B in his probability theory course, given that he attends TA sessions. To solve this, we will employ Bayes' Theorem, which is often used to find conditional probabilities.

Step 1: Define the Events

Let:

• $A$: Chang receives a grade higher than B.
• $D$: Chang attends the TA sessions.
• $D^c$: Chang does not attend the TA sessions.

From the problem, we know the following:

• $P(A^c | D^c) = 0.95$, meaning 95% of past students who did not attend TA sessions received a grade of B or below.
  • Therefore, $P(A | D^c) = 1 - P(A^c | D^c) = 1 - 0.95 = 0.05$, meaning 5% of students who did not attend TA sessions received a grade higher than B.
• $P(A) = 0.13$, meaning 13% of all students received a grade higher than B.
• $P(D) = \frac{9}{19}$, meaning 9 out of every 19 students attend TA sessions.

Step 2: Find $P(A | D)$

We are asked to compute the probability that Chang will receive a grade higher than B given that he attends TA sessions, $P(A | D)$. Using Bayes' Theorem, we can write:

$P(A | D) = \frac{P(D | A) P(A)}{P(D)}$

First, we need to find $P(D | A)$, the probability that a student attends TA sessions given they received a grade higher than B.

Step 3: Use the Law of Total Probability to Find $P(D)$

To find $P(D)$, we use the law of total probability, taking into account both students who received a grade higher than B and those who did not:

$P(D) = P(D | A) P(A) + P(D | A^c) P(A^c)$

But we do not yet know $P(D | A)$. Instead, we have enough information to calculate $P(A | D)$ using an alternate form of Bayes' Theorem.

Step 4: Bayes’ Theorem Calculation

Using the information we have:

• $P(A | D^c) = 0.05$
• $P(A) = 0.13$
• $P(D) = \frac{9}{19}$