The lecturer reported that 90% of the students who enrolled in statistics course passed the midterm test. A
tutorial class has 15 students.
(i) Find the probability that all students pass the midterm test. (2 marks)
(ii) Find the probability that more than two students failed the midterm test. (3 marks)

We are dealing with a binomial probability distribution in this scenario. The binomial distribution applies here because:

- Each student either passes or fails the midterm test (two possible outcomes).
- The probability of passing is constant for each student.
- The results of one student's test do not affect the others (independence).

Let:
- \( n = 15 \) (the number of students in the tutorial class),
- \( p = 0.90 \) (the probability of passing),
- \( q = 1 - p = 0.10 \) (the probability of failing).

### (i) Probability that all students pass the midterm test:

For all students to pass the test, all 15 students need to succeed. The probability of this happening is:

\[
P(\text{All pass}) = p^{15} = 0.90^{15}
\]

Let’s compute that.

\[
P(\text{All pass}) = (0.90)^{15} \approx 0.2059
\]

So, the probability that all students pass the test is approximately **0.2059** (or 20.59%).

### (ii) Probability that more than two students failed the midterm test:

We are asked to find the probability that more than two students failed, i.e., \( P(X > 2) \), where \( X \) is the number of students who failed. To calculate this, it's easier to find the probability that two or fewer students failed and subtract it from 1:

\[
P(X > 2) = 1 - P(X \leq 2)
\]

Where \( P(X \leq 2) \) represents the cumulative probability for 0, 1, or 2 students failing. We will use the binomial probability formula:

\[
P(X = k) = \binom{n}{k} \cdot p^{n-k} \cdot q^k
\]

We need to calculate the probabilities for \( k = 0 \), \( k = 1 \), and \( k = 2 \), then sum them up.

1. **For \( X = 0 \)** (no students fail):
   \[
   P(X = 0) = \binom{15}{0} \cdot (0.90)^{15} \cdot (0.10)^0 = 0.2059
   \]

2. **For \( X = 1 \)** (1 student fails):
   \[
   P(X = 1) = \binom{15}{1} \cdot (0.90)^{14} \cdot (0.10)^1 = 15 \cdot (0.90)^{14} \cdot 0.10
   \]

3. **For \( X = 2 \)** (2 students fail):
   \[
   P(X = 2) = \binom{15}{2} \cdot (0.90)^{13} \cdot (0.10)^2 = \frac{15 \cdot 14}{2} \cdot (0.90)^{13} \cdot 0.10^2
   \]

After finding these probabilities, sum them and subtract from 1 to find \( P(X > 2) \). Let me calculate that.### (i) The probability that all 15 students pass the midterm test is approximately **0.2059** (or 20.59%).

### (ii) The probability that more than two students fail the midterm test is approximately **0.1841** (or 18.41%).

Would you like more details on any of the steps?

Here are five related questions:
1. What is a binomial distribution, and how is it applied in probability problems?
2. How would the result change if the probability of passing was lower, such as 80%?
3. How do we interpret the result of the binomial coefficient in real-world scenarios?
4. What are other methods for calculating probabilities involving multiple trials?
5. How can cumulative probabilities be useful in different statistical scenarios?

**Tip**: In binomial distributions, calculating cumulative probabilities for a range of outcomes can simplify finding more complex probabilities like "more than" or "less than" scenarios.

The lecturer reported that 90% of the students who enrolled in statistics course passed the midterm test. A tutorial class has 15 students. (i) Find the probability that all students pass the midterm test. (ii) Find the probability that more than two students failed the midterm test.

Solve a statistics problem involving binomial distribution where 90% of students pass the midterm test in a class of 15. Find the probability that all students pass and that more than two students fail using binomial and cumulative probability formulas. Suitable for college-level statistics students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation