For a Binomial distribution \( B(n, p) \), \( \frac{E(x)}{V(x)} \) is equal to:
\( \frac{1}{1 - p} \)
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.
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:
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 \( V(x) = npq \).
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:
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 \( np \):
Thus, the ratio \( \frac{E(x)}{V(x)} \) for a Binomial distribution \( B(n, p) \) is equal to \( \frac{1}{1 - p} \).
Let's compare our calculated ratio \( \frac{1}{1 - p} \) with the given options:
Our derived ratio matches the fourth option.
| 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 |
A random variable follows a Binomial distribution if the following conditions are met:
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.
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?
Let \( X \) be a random variable whose probability distribution is given by the table:
| X | 1 | 3 | 5 | 7 |
|---|---|---|---|---|
| P(X) | \( \frac{1}{3} \) | \( \frac{1}{6} \) | \( \frac{1}{6} \) | \( \frac{1}{3} \) |
Then variance of \( X \) is:
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