The mean and variance of a Binomial Variate X with parameters n and p are 16 and 0.8
respectively. Find (i) P(X = 0) (ii) P(X ≥ 2).

Given that \( X \) is a binomial random variable with parameters \( n \) (number of trials) and \( p \) (probability of success), the mean (\(\mu\)) and variance (\(\sigma^2\)) of \( X \) are given by:

\[
\text{Mean: } \mu = np = 16
\]
\[
\text{Variance: } \sigma^2 = np(1-p) = 0.8
\]

### Step 1: Determine \( n \) and \( p \)
From the mean equation:

\[
np = 16 \quad \text{(Equation 1)}
\]

From the variance equation:

\[
np(1-p) = 0.8 \quad \text{(Equation 2)}
\]

Substitute \( np = 16 \) into Equation 2:

\[
16(1-p) = 0.8
\]

Solve for \( p \):

\[
1 - p = \frac{0.8}{16} = 0.05
\]
\[
p = 1 - 0.05 = 0.95
\]

Now, substitute \( p = 0.95 \) into Equation 1 to find \( n \):

\[
n(0.95) = 16
\]
\[
n = \frac{16}{0.95} \approx 16.84
\]

Since \( n \) must be an integer (as it's the number of trials), we round \( n \) to 17.

### Step 2: Find the required probabilities

(i) **\( P(X = 0) \)**

\[
P(X = 0) = \binom{17}{0} (0.95)^0 (0.05)^{17} = (0.05)^{17}
\]

\[
P(X = 0) = 0.05^{17} \approx 7.62 \times 10^{-22}
\]

(ii) **\( P(X \geq 2) \)**

\[
P(X \geq 2) = 1 - P(X = 0) - P(X = 1)
\]

First, calculate \( P(X = 1) \):

\[
P(X = 1) = \binom{17}{1} (0.95)^1 (0.05)^{16}
\]
\[
P(X = 1) = 17 \times 0.95 \times (0.05)^{16} \approx 1.23 \times 10^{-20}
\]

Now, calculate \( P(X \geq 2) \):

\[
P(X \geq 2) \approx 1 - 7.62 \times 10^{-22} - 1.23 \times 10^{-20} \approx 1
\]

So, \( P(X \geq 2) \approx 1 \).

### Final Answers:
1. \( P(X = 0) \approx 7.62 \times 10^{-22} \)
2. \( P(X \geq 2) \approx 1 \)

Do you want details on any specific steps or have any questions?

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Here are five related questions to consider:

1. How would you compute the probability \( P(X = 2) \) for the same binomial distribution?
2. What would be the effect on the mean and variance if the value of \( p \) was 0.9 instead of 0.95?
3. How would the probabilities change if \( n \) was a different integer, say 20?
4. What is the cumulative probability \( P(X \leq 3) \)?
5. How does the binomial distribution behave as \( n \) increases with a fixed \( p \)?

**Tip:** In binomial distributions, the variance is maximized when \( p = 0.5 \), and it decreases as \( p \) approaches 0 or 1.

Explore probability problems involving the binomial distribution with specific mean and variance. Learn how to calculate probabilities such as P(X = 0) and P(X ≥ 2) using formulas and step-by-step solutions.

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