For an initial screening of an admission test, a candidate is given fifty problems to solve. If the probability that the candidate can solve any proglem is `(4)/(5)`, then the probability that he is unable to solve less than two problem is

Options

A. `(201)/(5) ((1)/(5))^(49)`
B. `(316)/(25) ((4)/(5))^(48)`
C. `(54)/(5) ((4)/(5))^(49)`
D. `(164)/(25) ((1)/(5))^(48)`

Answer

To solve the problem, we need to find the probability that a candidate is unable to solve less than two problems out of fifty. This means we need to calculate the probability of the candidate solving 49 or 50 problems.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Total number of problems (N) = 50
- Probability of solving a problem (p) = 4/5
- Probability of not solving a problem (q) = 1 - p = 1/5

2. **Define the Required Probability:**
- We need to find the probability that the candidate is unable to solve less than two problems, which means he can solve either 49 or 50 problems.
- This can be expressed as:
\[ P(X \geq 49) = P(X = 49) + P(X = 50) \]

3. **Use the Binomial Probability Formula:**
- The binomial probability formula is given by:
\[ P(X = r) = \binom{N}{r} p^r q^{N-r} \]
- Here, \( \binom{N}{r} \) is the binomial coefficient, which represents the number of ways to choose r successes in N trials.

4. **Calculate \( P(X = 49) \):**
- For \( r = 49 \):
\[ P(X = 49) = \binom{50}{49} \left(\frac{4}{5}\right)^{49} \left(\frac{1}{5}\right)^{1} \]
- Calculate \( \binom{50}{49} = 50 \):
\[ P(X = 49) = 50 \left(\frac{4}{5}\right)^{49} \left(\frac{1}{5}\right) \]
- This simplifies to:
\[ P(X = 49) = 50 \cdot \frac{4^{49}}{5^{49}} \cdot \frac{1}{5} = \frac{50 \cdot 4^{49}}{5^{50}} \]

5. **Calculate \( P(X = 50) \):**
- For \( r = 50 \):
\[ P(X = 50) = \binom{50}{50} \left(\frac{4}{5}\right)^{50} \left(\frac{1}{5}\right)^{0} \]
- Calculate \( \binom{50}{50} = 1 \):
\[ P(X = 50) = 1 \cdot \left(\frac{4}{5}\right)^{50} \]
- This simplifies to:
\[ P(X = 50) = \frac{4^{50}}{5^{50}} \]

6. **Combine the Probabilities:**
- Now, we can combine both probabilities:
\[ P(X \geq 49) = P(X = 49) + P(X = 50) \]
- Substitute the values:
\[ P(X \geq 49) = \frac{50 \cdot 4^{49}}{5^{50}} + \frac{4^{50}}{5^{50}} \]
- Factor out \( \frac{4^{49}}{5^{50}} \):
\[ P(X \geq 49) = \frac{4^{49}}{5^{50}} \left(50 + 4\right) = \frac{4^{49}}{5^{50}} \cdot 54 \]

7. **Final Probability:**
- Therefore, the final probability that the candidate is unable to solve less than two problems is:
\[ P(X \geq 49) = \frac{54 \cdot 4^{49}}{5^{50}} \]