A multiple choice emamination has 5 questions. Each question has three alternative answers of which exactly one is correct. The probability that a student will get 4 or more correct answers just guessing is

Options

A. `(17)/(3^(5))`
B. `(13)/(3^(5))`
C. `(11)/(3^(5))`
D. `(10)/(3^(5))`

Explanation

To solve the problem, we need to calculate the probability that a student will get 4 or more correct answers by guessing on a multiple-choice exam with 5 questions, each having 3 alternatives (only one of which is correct).

### Step-by-Step Solution:

1. **Identify the Probability of Correct and Incorrect Answers:**
- The probability of guessing the correct answer for any question is \( P(\text{Correct}) = \frac{1}{3} \).
- The probability of guessing an incorrect answer is \( P(\text{Incorrect}) = \frac{2}{3} \).

2. **Define the Random Variable:**
- Let \( X \) be the random variable representing the number of correct answers. \( X \) follows a binomial distribution since each question is independent.
- The parameters for the binomial distribution are \( n = 5 \) (the number of questions) and \( p = \frac{1}{3} \) (the probability of a correct answer).

3. **Calculate the Required Probability:**
- We need to find \( P(X \geq 4) \), which can be expressed as:
\[ P(X \geq 4) = P(X = 4) + P(X = 5) \]

4. **Calculate \( P(X = 4) \):**
- The formula for the binomial probability is:
\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]
- For \( k = 4 \):
\[ P(X = 4) = \binom{5}{4} \left(\frac{1}{3}\right)^4 \left(\frac{2}{3}\right)^{5-4} \]
\[ = 5 \cdot \left(\frac{1}{3}\right)^4 \cdot \left(\frac{2}{3}\right)^1 \]
\[ = 5 \cdot \frac{1}{81} \cdot \frac{2}{3} = 5 \cdot \frac{2}{243} = \frac{10}{243} \]

5. **Calculate \( P(X = 5) \):**
- For \( k = 5 \):
\[ P(X = 5) = \binom{5}{5} \left(\frac{1}{3}\right)^5 \left(\frac{2}{3}\right)^{5-5} \]
\[ = 1 \cdot \left(\frac{1}{3}\right)^5 \cdot 1 = \frac{1}{243} \]

6. **Combine the Probabilities:**
- Now, we can sum the probabilities:
\[ P(X \geq 4) = P(X = 4) + P(X = 5) = \frac{10}{243} + \frac{1}{243} = \frac{11}{243} \]

### Final Answer:
The probability that a student will get 4 or more correct answers just by guessing is \( \frac{11}{243} \).