In this exercise, we are dealing with a sample space related to the outcomes of student evaluations in a Physics course. Each student can either pass (L) or fail (T). The sample space is described as:

$S = \{LLL, LLT, LTT, TTT, TLL, TLT, LTL, TTL\}$

We are given several events:

• K: No student fails (i.e., all pass). $K = \{LLL\}$

• L: At least one student fails. $L = \{LLT, LTT, TTT, TLL, TLT, LTL, TTL\}$

• M: Exactly one student fails. $M = \{LLT, LTL, TLL\}$

• N: Exactly two students fail. $N = \{LTT, TLT, LTL\}$

• O: One or two students fail. $O = \{LLT, LTL, LTT, TLT, TTL\}$

Question 1: Tentukan peluang kejadian K, L, M, N, O

To calculate the probability of each event, we will use the formula:

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

Where:

• $|A|$ is the number of outcomes in event $A$.
• $|S|$ is the total number of outcomes in the sample space $S$, which in this case is $|S| = 8$.

Let's calculate the probabilities of each event $K, L, M, N, O$.The probabilities for each event are as follows:

• $P(K) = 0.125$ (No student fails)
• $P(L) = 0.875$ (At least one student fails)
• $P(M) = 0.375$ (Exactly one student fails)
• $P(N) = 0.375$ (Exactly two students fail)
• $P(O) = 0.625$ (One or two students fail)

Do you have any questions or would you like more details on this?

Here are 5 follow-up questions based on this problem:

1. How would you determine the probability if there were more students?
2. What is the probability that exactly three students fail?
3. How would the probabilities change if the number of students passing was more likely than failing?
4. Can you calculate the complement of event $L$ and verify the result?
5. How would you approach this problem using a probability tree?

Tip: The sum of probabilities for all possible outcomes in a sample space should always equal 1.