On a multiple choice examination with three possible answers (out of which only one is correct) for each of the five questions, what is the probability that a candidate would get four or more correct answers just by guessing.

Answer

To solve the problem of finding the probability that a candidate would get four or more correct answers just by guessing on a multiple-choice examination with three possible answers for each of the five questions, we can follow these steps:

### Step 1: Define the Problem
We have a multiple-choice exam with 5 questions, each having 3 possible answers (only one of which is correct). We need to find the probability of getting at least 4 correct answers by guessing.

### Step 2: Identify the Distribution
Since the candidate is guessing, the number of correct answers follows a binomial distribution. Let:
- \( n = 5 \) (the number of questions),
- \( p = \frac{1}{3} \) (the probability of guessing a question correctly),
- \( q = 1 - p = \frac{2}{3} \) (the probability of guessing incorrectly).

### Step 3: Write the Binomial Probability Formula
The probability of getting exactly \( k \) correct answers in \( n \) trials is given by the binomial probability formula:
\[ P(X = k) = \binom{n}{k} p^k q^{n-k} \]
where \( \binom{n}{k} \) is the binomial coefficient.

### Step 4: Calculate the Probability for 4 and 5 Correct Answers
We need to calculate \( P(X \geq 4) = P(X = 4) + P(X = 5) \).

#### For \( P(X = 4) \):
\[ P(X = 4) = \binom{5}{4} \left(\frac{1}{3}\right)^4 \left(\frac{2}{3}\right)^{5-4} \]
Calculating this:
\[ \binom{5}{4} = 5 \]
\[ P(X = 4) = 5 \cdot \left(\frac{1}{3}\right)^4 \cdot \left(\frac{2}{3}\right)^1 = 5 \cdot \frac{1}{81} \cdot \frac{2}{3} = 5 \cdot \frac{2}{243} = \frac{10}{243} \]

#### For \( P(X = 5) \):
\[ P(X = 5) = \binom{5}{5} \left(\frac{1}{3}\right)^5 \left(\frac{2}{3}\right)^{5-5} \]
Calculating this:
\[ \binom{5}{5} = 1 \]
\[ P(X = 5) = 1 \cdot \left(\frac{1}{3}\right)^5 \cdot 1 = \frac{1}{243} \]

### Step 5: Combine the Probabilities
Now, we combine the probabilities for 4 and 5 correct answers:
\[ P(X \geq 4) = P(X = 4) + P(X = 5) = \frac{10}{243} + \frac{1}{243} = \frac{11}{243} \]

### Final Answer
The probability that a candidate would get four or more correct answers just by guessing is:
\[ \frac{11}{243} \]