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Question

For a Binomial distribution \( B(n, p) \), \( \frac{E(x)}{V(x)} \) is equal to:

The correct answer is

\( \frac{1}{1 - p} \)

Understanding Binomial Distribution Properties

A Binomial distribution \( B(n, p) \) is a discrete probability distribution that represents the number of successes in a fixed number of independent Bernoulli trials, where \( n \) is the number of trials and \( p \) is the probability of success on a single trial.

Calculating Expected Value and Variance

For a Binomial distribution \( B(n, p) \), the expected value, often denoted as \( E(x) \) or \( \mu \), is the average outcome expected over many trials. The variance, denoted as \( V(x) \) or \( \sigma^2 \), measures the spread or dispersion of the distribution.

The formulas for the expected value and variance of a Binomial distribution \( B(n, p) \) are well-established:

  • Expected Value \( E(x) = np \)
  • Variance \( V(x) = np(1 - p) \)

Here, \( n \) is the number of trials and \( p \) is the probability of success in a single trial. The term \( (1 - p) \) is often denoted as \( q \), representing the probability of failure, so \( V(x) = npq \).

Determining the Ratio \( \frac{E(x)}{V(x)} \)

The question asks for the ratio of the expected value to the variance for a Binomial distribution \( B(n, p) \). We can find this ratio by dividing the formula for \( E(x) \) by the formula for \( V(x) \).

The ratio is:

\( \frac{E(x)}{V(x)} = \frac{np}{np(1 - p)} \)

Assuming \( n > 0 \) and \( p > 0 \) (for a non-degenerate distribution), the term \( np \) appears in both the numerator and the denominator. We can cancel out \( np \):

\( \frac{np}{np(1 - p)} = \frac{1}{1 - p} \)

Thus, the ratio \( \frac{E(x)}{V(x)} \) for a Binomial distribution \( B(n, p) \) is equal to \( \frac{1}{1 - p} \).

Comparing with Options

Let's compare our calculated ratio \( \frac{1}{1 - p} \) with the given options:

  1. \( \frac{1}{p^2} \)
  2. \( 1 - p \)
  3. \( \frac{1 + p}{1 + p} = 1 \)
  4. \( \frac{1}{1 - p} \)

Our derived ratio matches the fourth option.

Revision Table: Binomial Distribution Formulas

Property Formula for \( B(n, p) \) Notes
Probability Mass Function \( P(X=k) \) \( \binom{n}{k} p^k (1-p)^{n-k} \) For \( k = 0, 1, ..., n \)
Expected Value \( E(x) \) \( np \) Mean of the distribution
Variance \( V(x) \) \( np(1-p) \) Square of the standard deviation
Standard Deviation \( \sigma \) \( \sqrt{np(1-p)} \) Spread of the distribution

Additional Information: Binomial Distribution Conditions

A random variable follows a Binomial distribution if the following conditions are met:

  • There are a fixed number of trials, denoted by \( n \).
  • Each trial is independent of the others.
  • There are only two possible outcomes for each trial: success (with probability \( p \)) and failure (with probability \( 1-p \)).
  • The probability of success \( p \) is the same for each trial.

The ratio \( \frac{E(x)}{V(x)} = \frac{1}{1 - p} \) shows how the mean relates to the spread, scaled by the probability of failure. As \( p \) approaches 1 (success is very likely), \( 1-p \) approaches 0, and the ratio \( \frac{1}{1-p} \) becomes very large. This indicates that the mean is much larger than the variance when success is highly probable, which makes sense as the distribution becomes heavily skewed towards \( n \) successes with little spread.

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Similar Questions

  1. Which of the following is the probability of \( x \) successes in a binomial distribution with number of trials \( n \) and probability of success as \( \theta \) ( \( 0 < \theta < 1 \) ) in each trial?

  2. Let \( X \) be a random variable whose probability distribution is given by the table:

    X1357
    P(X)\( \frac{1}{3} \)\( \frac{1}{6} \)\( \frac{1}{6} \)\( \frac{1}{3} \)

    Then variance of \( X \) is:

  3. The probability that the problem is solved is:

  4. The probability that either only Kalyani or Kashvi or Sara solves it is:


Important Questions from Probability

  1. A die is rolled thrice. What is the probability of getting a number greater than 4 in the first and the second throws, and a number less than 4 in the third throw?

  2. Two dice are thrown simultaneously. If \( X \) denotes the number of fours, then the expectation of \( X \) will be:

  3. If the random variable \( X \) has the following distribution:

    X012otherwise
    P(X)k2k3k0

     

     

    Match List-I with List-II:

    List-IList-II
    (A) k(I) \(\frac{5}{6}\)
    (B) P(X < 2)(II) \(\frac{4}{3}\)
    (C) E(X)(III) \(\frac{1}{2}\)
    (D) P(1 ≤ X ≤ 2)(IV) \(\frac{1}{6}\)

    Choose the correct answer from the options given below:

     

  4. Let X denote the number of hours you play during a randomly selected day. The probability that X can take values x has the following form, where c is some constant:

    \[ P(X = x) = \begin{cases} 0.1, & \text{if } x = 0 \\ cx, & \text{if } x = 1 \text{ or } x = 2 \\ c(5 - x), & \text{if } x = 3 \text{ or } x = 4 \\ 0, & \text{otherwise} \end{cases} \]

     

     

           

    Match List-I with List-II:

    List-IList-II
    (A) c(I) 0.75
    (B) P(X ≤ 2)(II) 0.3
    (C) P(X = 2)(III) 0.55
    (D) P(X ≥ 2)(IV) 0.15

    Choose the correct answer from the options given below:

     

  5. For the differential equation \( (x \log_e x) dy = (\log_e x - y) dx \):

    (A) Degree of the given differential equation is 1.

    (B) It is a homogeneous differential equation.

    (C) Solution is \( 2y \log_e x + A = (\log_e x)^2 \), where A is an arbitrary constant.

    (D) Solution is \( 2y \log_e x + A = \log_e (\log_e x) \), where A is an arbitrary constant.

    Choose the correct answer from the options given below:

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