The probability of a student clearing a competitive exam is 
1
4
4
1
​
 . If he gives the exam 9 times, then what is the probability of him clearing the exam at least twice?(Enter the answer correct to 2 decimal places)

To solve this problem, we need to determine the probability that a student clears the exam at least twice out of 9 attempts. This can be modeled using the binomial distribution.

### Step 1: Define the parameters
- Probability of success (clearing the exam) on a single attempt: \( p = \frac{1}{4} \).
- Number of trials (attempts): \( n = 9 \).
- We want to find the probability of the student clearing the exam at least twice, which means we need to calculate \( P(X \geq 2) \), where \( X \) is the number of exams cleared.

### Step 2: Use the binomial probability formula
The probability mass function of a binomial distribution is given by:
\[
P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}
\]
where \( \binom{n}{k} \) is the binomial coefficient.

### Step 3: Calculate \( P(X \geq 2) \)
The probability of clearing the exam at least twice is:
\[
P(X \geq 2) = 1 - P(X < 2)
\]
This can be expanded as:
\[
P(X \geq 2) = 1 - \left[ P(X = 0) + P(X = 1) \right]
\]

### Step 4: Calculate \( P(X = 0) \) and \( P(X = 1) \)

1. **For \( X = 0 \):**
\[
P(X = 0) = \binom{9}{0} \left(\frac{1}{4}\right)^0 \left(\frac{3}{4}\right)^9 = 1 \times \left(\frac{3}{4}\right)^9
\]
\[
P(X = 0) = \left(\frac{3}{4}\right)^9 \approx 0.0751
\]

2. **For \( X = 1 \):**
\[
P(X = 1) = \binom{9}{1} \left(\frac{1}{4}\right)^1 \left(\frac{3}{4}\right)^8 = 9 \times \frac{1}{4} \times \left(\frac{3}{4}\right)^8
\]
\[
P(X = 1) = 9 \times \frac{1}{4} \times 0.1001 \approx 0.2252
\]

### Step 5: Calculate \( P(X \geq 2) \)
\[
P(X \geq 2) = 1 - \left(0.0751 + 0.2252\right) \approx 1 - 0.3003 = 0.6997
\]

Thus, the probability that the student clears the exam at least twice out of 9 attempts is approximately **0.70** (correct to 2 decimal places).

Would you like more details or have any questions? Here are some related questions:

1. How would the probability change if the student attempts the exam more times?
2. What is the probability of the student clearing the exam exactly 3 times out of 9?
3. How does the probability change if the success rate improves to \( \frac{1}{3} \)?
4. Can we use the Poisson approximation to solve this problem?
5. What would be the expected number of successful attempts out of 9?

**Tip:** When dealing with binomial probabilities for "at least" cases, it is often easier to calculate the complementary probability (i.e., the probability of fewer successes) and subtract it from 1.

Explore how to calculate the probability of a student clearing an exam at least twice out of 9 attempts using binomial distribution. Understand the binomial probability formula and its application. Perfect for high school students studying probability and statistics.

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