In an examination, the probability of a candidate solving a question is `(1)/(2)`. Out of given 5 questions in the examination, what is the probability that the candidate was able to solve at least 2 question?

Options

A. `(1)/(64)`
B. `(3)/(16)`
C. `(1)/(2)`
D. `(13)/(16)`

Answer

To find the probability that a candidate solves at least 2 questions out of 5, where the probability of solving each question is \( \frac{1}{2} \), we can follow these steps:

### Step 1: Define the problem
We need to calculate the probability of the candidate solving at least 2 questions. This means we need to find the probabilities of the candidate solving 2, 3, 4, or 5 questions.

### Step 2: Use the binomial probability formula
The probability of solving exactly \( k \) questions out of \( n \) total questions can be calculated using the binomial probability formula:
\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]
where:
- \( n \) = total number of questions (5 in this case)
- \( k \) = number of questions solved
- \( p \) = probability of solving a question (\( \frac{1}{2} \))
- \( \binom{n}{k} \) = combination of \( n \) questions taken \( k \) at a time

### Step 3: Calculate probabilities for \( k = 2, 3, 4, 5 \)
We will calculate the probabilities for \( k = 2, 3, 4, 5 \):

1. **For \( k = 2 \)**:
\[ P(X = 2) = \binom{5}{2} \left(\frac{1}{2}\right)^2 \left(\frac{1}{2}\right)^{5-2} = \binom{5}{2} \left(\frac{1}{2}\right)^5 \]
\[ = 10 \cdot \frac{1}{32} = \frac{10}{32} \]

2. **For \( k = 3 \)**:
\[ P(X = 3) = \binom{5}{3} \left(\frac{1}{2}\right)^3 \left(\frac{1}{2}\right)^{5-3} = \binom{5}{3} \left(\frac{1}{2}\right)^5 \]
\[ = 10 \cdot \frac{1}{32} = \frac{10}{32} \]

3. **For \( k = 4 \)**:
\[ P(X = 4) = \binom{5}{4} \left(\frac{1}{2}\right)^4 \left(\frac{1}{2}\right)^{5-4} = \binom{5}{4} \left(\frac{1}{2}\right)^5 \]
\[ = 5 \cdot \frac{1}{32} = \frac{5}{32} \]

4. **For \( k = 5 \)**:
\[ P(X = 5) = \binom{5}{5} \left(\frac{1}{2}\right)^5 \left(\frac{1}{2}\right)^{5-5} = \binom{5}{5} \left(\frac{1}{2}\right)^5 \]
\[ = 1 \cdot \frac{1}{32} = \frac{1}{32} \]

### Step 4: Sum the probabilities
Now, we sum the probabilities for \( k = 2, 3, 4, 5 \):
\[ P(X \geq 2) = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) \]
\[ = \frac{10}{32} + \frac{10}{32} + \frac{5}{32} + \frac{1}{32} = \frac{26}{32} \]

### Step 5: Simplify the result
\[ \frac{26}{32} = \frac{13}{16} \]

### Final Answer
The probability that the candidate was able to solve at least 2 questions is \( \frac{13}{16} \).

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