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# Geometric Distribution Formula

• Difficulty Level : Easy
• Last Updated : 04 Apr, 2022

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-1p

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 -1p

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) / p2

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

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 – 1p

P(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 – 1p

P(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, 3

P(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/ p2

= 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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