Math Problem Statement

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).

Solution

Given that XX is a binomial random variable with parameters nn (number of trials) and pp (probability of success), the mean (μ\mu) and variance (σ2\sigma^2) of XX are given by:

Mean: μ=np=16\text{Mean: } \mu = np = 16 Variance: σ2=np(1p)=0.8\text{Variance: } \sigma^2 = np(1-p) = 0.8

Step 1: Determine nn and pp

From the mean equation:

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

From the variance equation:

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

Substitute np=16np = 16 into Equation 2:

16(1p)=0.816(1-p) = 0.8

Solve for pp:

1p=0.816=0.051 - p = \frac{0.8}{16} = 0.05 p=10.05=0.95p = 1 - 0.05 = 0.95

Now, substitute p=0.95p = 0.95 into Equation 1 to find nn:

n(0.95)=16n(0.95) = 16 n=160.9516.84n = \frac{16}{0.95} \approx 16.84

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

Step 2: Find the required probabilities

(i) P(X=0)P(X = 0)

P(X=0)=(170)(0.95)0(0.05)17=(0.05)17P(X = 0) = \binom{17}{0} (0.95)^0 (0.05)^{17} = (0.05)^{17}

P(X=0)=0.05177.62×1022P(X = 0) = 0.05^{17} \approx 7.62 \times 10^{-22}

(ii) P(X2)P(X \geq 2)

P(X2)=1P(X=0)P(X=1)P(X \geq 2) = 1 - P(X = 0) - P(X = 1)

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

P(X=1)=(171)(0.95)1(0.05)16P(X = 1) = \binom{17}{1} (0.95)^1 (0.05)^{16} P(X=1)=17×0.95×(0.05)161.23×1020P(X = 1) = 17 \times 0.95 \times (0.05)^{16} \approx 1.23 \times 10^{-20}

Now, calculate P(X2)P(X \geq 2):

P(X2)17.62×10221.23×10201P(X \geq 2) \approx 1 - 7.62 \times 10^{-22} - 1.23 \times 10^{-20} \approx 1

So, P(X2)1P(X \geq 2) \approx 1.

Final Answers:

  1. P(X=0)7.62×1022P(X = 0) \approx 7.62 \times 10^{-22}
  2. P(X2)1P(X \geq 2) \approx 1

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


Here are five related questions to consider:

  1. How would you compute the probability P(X=2)P(X = 2) for the same binomial distribution?
  2. What would be the effect on the mean and variance if the value of pp was 0.9 instead of 0.95?
  3. How would the probabilities change if nn was a different integer, say 20?
  4. What is the cumulative probability P(X3)P(X \leq 3)?
  5. How does the binomial distribution behave as nn increases with a fixed pp?

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

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Math Problem Analysis

Mathematical Concepts

Binomial Distribution
Probability

Formulas

Binomial probability formula
Mean and variance of binomial distribution

Theorems

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Suitable Grade Level

Advanced High School