Suppose a random variable, *x*, arises from a binomial experiment. If *n* = 10, and *p* = 0.1, find the following probabilities using the binomial formula. Round to four decimal places, if necessary.

P(x = 9) = 

P(x = 5) = 

P(x = 2) = 

P(x ≤ 1) = 

P(x ≥ 8) = 

P(x ≤ 7) = 

To solve these binomial probabilities, we will use the binomial probability formula:

\[
P(x = k) = \binom{n}{k} p^k (1 - p)^{n - k}
\]

where:
- \( n = 10 \) (the number of trials),
- \( p = 0.1 \) (the probability of success),
- \( 1 - p = 0.9 \) (the probability of failure),
- \( k \) is the number of successes we want to find the probability for,
- \(\binom{n}{k}\) is the binomial coefficient, calculated as \( \frac{n!}{k!(n-k)!} \).

Now, we calculate each of the requested probabilities:

### 1. \( P(x = 9) \):
\[
P(x = 9) = \binom{10}{9} (0.1)^9 (0.9)^1
= \frac{10!}{9!(10-9)!} (0.1)^9 (0.9)
= 10 \cdot (0.1)^9 \cdot 0.9
\]
\[
P(x = 9) = 10 \cdot 0.000000001 \cdot 0.9 = 0.000000009
\]
Rounded to four decimal places: \( P(x = 9) = 0.0000 \).

### 2. \( P(x = 5) \):
\[
P(x = 5) = \binom{10}{5} (0.1)^5 (0.9)^5
= \frac{10!}{5!(10-5)!} (0.1)^5 (0.9)^5
= 252 \cdot (0.1)^5 \cdot (0.9)^5
\]
\[
P(x = 5) = 252 \cdot 0.00001 \cdot 0.59049 = 0.001488
\]
So, \( P(x = 5) = 0.0015 \).

### 3. \( P(x = 2) \):
\[
P(x = 2) = \binom{10}{2} (0.1)^2 (0.9)^8
= \frac{10!}{2!(10-2)!} (0.1)^2 (0.9)^8
= 45 \cdot (0.1)^2 \cdot (0.9)^8
\]
\[
P(x = 2) = 45 \cdot 0.01 \cdot 0.43046721 = 0.19371
\]
So, \( P(x = 2) = 0.1937 \).

### 4. \( P(x \leq 1) \):
This represents the cumulative probability of \( P(x = 0) \) and \( P(x = 1) \).

#### \( P(x = 0) \):
\[
P(x = 0) = \binom{10}{0} (0.1)^0 (0.9)^{10}
= 1 \cdot 1 \cdot (0.9)^{10} = 0.3487
\]

#### \( P(x = 1) \):
\[
P(x = 1) = \binom{10}{1} (0.1)^1 (0.9)^9
= 10 \cdot 0.1 \cdot (0.9)^9 = 0.3874
\]

Now, sum the two probabilities:
\[
P(x \leq 1) = P(x = 0) + P(x = 1) = 0.3487 + 0.3874 = 0.7361
\]

### 5. \( P(x \geq 8) \):
This represents the cumulative probability of \( P(x = 8) \), \( P(x = 9) \), and \( P(x = 10) \).

#### \( P(x = 8) \):
\[
P(x = 8) = \binom{10}{8} (0.1)^8 (0.9)^2 = 45 \cdot 0.00000001 \cdot 0.81 = 0.00000036
\]
#### \( P(x = 10) \):
\[
P(x = 10) = \binom{10}{10} (0.1)^{10} (0.9)^0 = 1 \cdot 0.0000000001 = 0.0000000001
\]

Now, sum the probabilities for \( x = 8 \), \( x = 9 \), and \( x = 10 \):
\[
P(x \geq 8) = 0.00000036 + 0.000000009 + 0.0000000001 = 0.0000003691
\]
So, rounded to four decimal places: \( P(x \geq 8) = 0.0000 \).

### 6. \( P(x \leq 7) \):
This is the complement of \( P(x \geq 8) \):
\[
P(x \leq 7) = 1 - P(x \geq 8) = 1 - 0.0000 = 1.0000
\]

### Final Answers:
1. \( P(x = 9) = 0.0000 \)
2. \( P(x = 5) = 0.0015 \)
3. \( P(x = 2) = 0.1937 \)
4. \( P(x \leq 1) = 0.7361 \)
5. \( P(x \geq 8) = 0.0000 \)
6. \( P(x \leq 7) = 1.0000 \)

Would you like more details or have any other questions?

Here are 5 related questions you might explore:
1. What is the expected value and variance for this binomial distribution?
2. How does increasing \(p\) affect the shape of the binomial distribution?
3. Can you calculate \( P(2 \leq x \leq 5) \) for this distribution?
4. What happens to the probabilities if \(n\) is increased to 20?
5. How would you approximate these probabilities using a normal distribution?

**Tip**: The binomial distribution can be approximated by a normal distribution when \(n\) is large and \(p\) is not too close to 0 or 1.

Suppose a random variable, *x*, arises from a binomial experiment. If n = 10, and p = 0.1, find the following probabilities using the binomial formula. Round to four decimal places, if necessary.

P(x = 9) = 
P(x = 5) = 
P(x = 2) = 
P(x ≤ 1) = 
P(x ≥ 8) = 
P(x ≤ 7) =

Calculate binomial probabilities for a distribution where n = 10 and p = 0.1. Learn step-by-step how to find P(x = 9), P(x = 5), P(x = 2), P(x ≤ 1), P(x ≥ 8), and P(x ≤ 7), using the binomial formula. Suitable for high school students in Grades 9-12.

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