# Geometric Distribution Formula

In a Bernoulli trial, the likelihood of the number of successive failures before success is obtained is represented by a **geometric distribution**, which is a sort of discrete probability distribution. A Bernoulli trial is a test that can only have one of two outcomes: success or failure. In other words, a Bernoulli trial is repeated until success is obtained and then stopped in geometric distribution.

In a variety of real-life circumstances, the geometric distribution is commonly used. In the financial industry, for example, the geometric distribution is used to estimate the financial rewards of making a given decision in a cost-benefit analysis. In this post, we’ll look at the definition of the geometric distribution, several instances, and some associated topics.

### Geometric Distribution

A geometric distribution is a discrete probability distribution that indicates the likelihood of achieving one’s first success after a series of failures. The number of attempts in a geometric distribution can go on indefinitely until the first success is achieved. Geometric distributions are probability distributions that are based on three key assumptions.

- The trials that are being undertaken are self-contained.
- Each trial may only have one of two outcomes: success or failure.
- For each trial, the success probability, represented by p, is the same.

### Formula for Geometric Distribution

P (X = x) = (1-p)^{x-1}p

P (X â‰¤ x) = 1-(1-p)^{x}

The probability mass function (pmf) and the cumulative distribution function can both be used to characterize a geometric distribution (CDF). The chance of a trial’s success is denoted by p, whereas the likelihood of failure is denoted by q. q = 1 – p in this case. X âˆ¼ G ( p ) represents a discrete random variable, X, with a geometric probability distribution.

**Geometric Distribution PMF**

The likelihood that a discrete random variable, X, will be exactly identical to some value, x, is determined by the probability mass function.

**P(X = x) = (1 – p) ^{x -1}p**

where, 0 < p â‰¤ 1.

**Geometric Distribution CDF**

The probability that a random variable, X, will assume a value that is less than or equal to x can be described as the cumulative distribution function of a random variable, X, that is assessed at a point, x. The distribution function is another name for it.

**P(X â‰¤ x) = 1 – (1 – p) ^{x}**

**Mean of Geometric Distribution**

The geometric distribution’s mean is also the geometric distribution’s expected value. The weighted average of all values of a random variable, X, is the expected value of X.

**E[X] = 1 / p**

**Variance of Geometric Distribution**

Variance is a measure of dispersion that examines how far data in distribution is spread out in relation to the mean.

**Var[X] = (1 – p) / p ^{2}**

**Standard Deviation of Geometric Distribution**

The square root of the variance can be used to calculate the standard deviation. The standard deviation also indicates how far the distribution deviates from the mean.

**S.D. = âˆšVAR[X]**

**S.D. = âˆš1-p / p**

### Sample Problems

**Problem 1: If a patient is waiting for a suitable blood donor and the probability that the selected donor will be a match is 0.2, then find the expected number of donors who will be tested till a match is found including the matched donor.**

**Solution:**

Given,

p = 0.2

E[X] = 1 / p

= 1 / 0.2

= 5

The expected number of donors who will be tested till a match is found is

5

**Problem 2: Suppose you are playing a game of darts. The probability of success is 0.4. What is the probability that you will hit the bullseye on the third try?**

**Solution:**

Given,

p = 0.4

P(X = x) = (1 – p)

^{x – 1}pP(X = 3) = (1 – 0.4)

^{3 – 1}(0.4)P(X = 3) = (0.6)

^{2}(0.4)= 0.144

The probability that you will hit the bullseye on the third try is

0.144

**Problem 3: A light bulb manufacturing factory finds 3 in every 60 light bulbs defective. What is the probability that** **the first defective light bulb with be found when the 6th one is tested?**

**Solution:**

Given,

p = 3 / 60 = 0.05

P(X = x) = (1 – p)

^{x – 1}pP(X = 6) = (1 – 0.05)

^{6 – 1}(0.05)P(X = 6) = (0.95)

^{5}(0.05)P(X = 6) = 0.0386

The probability that the first defective light bulb is found on the 6th trial is

0.0368

**Problem 4: Find the probability density of geometric distribution if the value of p is 0.42; x = 1,2,3 and also calculate the mean and variance.**

**Solution:**

Given that p = 0.42 and the value of x = 1, 2, 3

The formula of probability density of geometric distribution is

P(x) = p (1-p)

^{x-1}; x =1, 2, 3P(x) = 0; otherwise

P(x) = 0.42 (1- 0.42)

P(x) = 0; Otherwise

Mean= 1/p = 1/0.42 = 2.380

Variance = 1-p/ p

^{2}= 1-0.42 /(0.42)

^{2}=

3.287

**Problem 5: If the probability of breaking the pot in the pool is 0.4, find the number of brakes before success and the corresponding variance and standard deviation.**

**Solution:**

Here,

X âˆ¼ geo(0.4)

Hence,

e(x) = 1/0.4 = 2.5

Var(x) = 0.6/0.4Â²

= 3.75

Hence, standard deviation ( Ïƒ) =

1.94

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