Math Problem Statement

Let 0 < p < 1 and n = 1, 2, 3, ... A certain experiment has a probability p of success and a probability 1 - p of failure. The experiment is repeated n times, with all experiments independent. Let X be the number of successes among the n trials. Verify that the sum of probabilities of the binomial distribution equals 1, compute the expected value E[X], and compute the variance Var[X].

Solution

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Problem Outline:

Given an experiment with probability pp of success and 1p1 - p of failure, repeated nn times with independent trials, we are dealing with a binomial distribution. Let XX represent the number of successes among the nn trials.

The probability mass function (PMF) for XX is given by:

P(X=k)=(nk)pk(1p)nk,k=0,1,2,,n.P(X = k) = \binom{n}{k} p^k (1 - p)^{n-k}, \quad k = 0, 1, 2, \dots, n.

Random variables with this PMF are called binomial random variables with parameters nn and pp.

Part (a): Verify the following identity:

k=0nP(X=k)=1\sum_{k=0}^{n} P(X = k) = 1 This shows that the total probability across all possible outcomes kk must sum to 1. The hint suggests using the Binomial Theorem for verification.

Part (b): Compute the expected value E[X]E[X].

We are given a hint that: k(nk)=n(n1k1).k \binom{n}{k} = n \binom{n-1}{k-1}. This suggests we use the properties of binomial coefficients to simplify the computation of the expected value.

Part (c): Compute the variance Var[X]\text{Var}[X].

Step-by-step solutions:

(a) Verifying k=0nP(X=k)=1\sum_{k=0}^{n} P(X = k) = 1

We need to prove that the sum of the binomial probabilities across all kk from 0 to nn equals 1. Using the PMF of the binomial distribution:

k=0n(nk)pk(1p)nk.\sum_{k=0}^{n} \binom{n}{k} p^k (1 - p)^{n-k}.

This is exactly the Binomial Theorem, which states that:

(p+(1p))n=1n=1.(p + (1 - p))^n = 1^n = 1.

Thus, the sum of all binomial probabilities equals 1, as required.

(b) Compute E[X]E[X]

The expected value of a binomial random variable XX with parameters nn and pp is known to be:

E[X]=np.E[X] = n p.

To derive this, we use the property of linearity of expectation. Specifically, the binomial random variable can be seen as the sum of nn independent Bernoulli trials, where each trial has an expected value pp. Therefore:

E[X]=k=0nkP(X=k).E[X] = \sum_{k=0}^{n} k \cdot P(X = k).

Using the given hint that:

k(nk)=n(n1k1),k \binom{n}{k} = n \binom{n-1}{k-1},

we can simplify the sum, eventually arriving at:

E[X]=np.E[X] = n p.

(c) Compute Var[X]\text{Var}[X]

The variance of a binomial random variable XX is:

Var(X)=np(1p).\text{Var}(X) = n p (1 - p).

This result can be derived using the fact that Var(X)=E[X2](E[X])2\text{Var}(X) = E[X^2] - (E[X])^2, and utilizing the properties of binomial random variables.

Summary:

  • (a): The sum of binomial probabilities equals 1, by the Binomial Theorem.
  • (b): The expected value E[X]=npE[X] = n p.
  • (c): The variance Var[X]=np(1p)\text{Var}[X] = n p (1 - p).

Would you like a more detailed explanation for any of these parts?

Here are 5 follow-up questions related to this problem:

  1. How does the binomial theorem help simplify the sum in part (a)?
  2. What is the significance of the expected value in the context of binomial random variables?
  3. Can you derive the formula for variance using moment generating functions?
  4. How would the expected value and variance change if p=12p = \frac{1}{2}?
  5. How would you compute higher moments (like skewness or kurtosis) for the binomial distribution?

Tip: Remember that for binomial distributions, you can always interpret the expected value E[X]E[X] as the product of the number of trials and the probability of success per trial.

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Math Problem Analysis

Mathematical Concepts

Probability
Binomial Distribution
Expected Value
Variance

Formulas

P(X = k) = (n choose k) * p^k * (1 - p)^(n-k)
E[X] = n * p
Var(X) = n * p * (1 - p)

Theorems

Binomial Theorem

Suitable Grade Level

College level or advanced high school