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Last updated on September 24, 2025
In probability and statistics, the geometric distribution represents the number of trials needed to get the first success in a series of independent and identically distributed Bernoulli trials. In this topic, we will learn the formula for the geometric distribution, including its mean, variance, and properties.
The geometric distribution is used to model the number of trials until the first success in a Bernoulli process.
Let’s learn the formulas related to geometric distribution, including its probability mass function, mean, and variance.
The geometric distribution formula gives the probability that the first success occurs on the k-th trial. It is calculated using the formula:
Probability mass function: \(P(X = k) = (1-p)^{k-1} \cdot p \) where p is the probability of success on each trial, and k is the trial number.
The mean of a geometric distribution represents the expected number of trials until the first success.
The formula for the mean is: Mean = \(\frac{1}{p}\) where p is the probability of success on each trial.
The variance of a geometric distribution measures the variability of the number of trials needed to get the first success. The formula for the variance is:
Variance = \(\frac{1-p}{p^2} \) where p is the probability of success on each trial.
In probability and statistics, the geometric distribution formula is crucial for understanding processes with binary outcomes. Here are some important applications:
It helps model scenarios like the number of coin flips until the first heads or the number of trials to find a defective item in quality control.
Understanding the mean and variance helps predict the average and variability of trials needed for the first success.
Students might find probability formulas tricky, but these tips can help:
Remember that the geometric distribution models "first success" scenarios, with the mean formula as 1/p indicating average trials needed.
Link the formula to real-life situations, like guessing the number of times you need to roll a die to get a six.
Use simple mnemonic devices like "first success equals geometric" to remember the distribution's context.
Students might face challenges when using geometric distribution formulas. Here are some mistakes and how to avoid them:
If the probability of flipping heads on a coin is 0.25, what's the probability of getting the first heads on the 3rd flip?
The probability is 0.140625
Using the formula \( P(X = k) = (1-p)^{k-1} \cdot p\) :
\(P(X = 3) = (1-0.25)^{3-1} \cdot 0.25 = 0.75^2 \cdot 0.25 = 0.140625\) .
What is the expected number of times you need to roll a die to get a 6?
The expected number is 6
The probability of rolling a 6 is \( \frac{1}{6}\) .
The mean is \(\frac{1}{p} = \frac{1}{\frac{1}{6}} = 6\) .
A factory produces light bulbs with a 0.02 probability of being defective. On average, how many bulbs will you test to find the first defective one?
The average number is 50
The probability of finding a defective bulb is 0.02.
The mean is \( \frac{1}{0.02} = 50\) .
If the probability that a customer makes a purchase is 0.1, how many customers on average will be approached before the first purchase?
The average number is 10
The probability of a purchase is 0.1.
The mean is \(\frac{1}{0.1} = 10\) .
In a game, the probability of hitting a target is 0.4. What is the probability of hitting the target for the first time on the 5th attempt?
The probability is 0.07776
Using the formula \(P(X = k) = (1-p)^{k-1} \cdot p\) :
\(P(X = 5) = (1-0.4)^{5-1} \cdot 0.4 = 0.6^4 \cdot 0.4 = 0.07776\) .
Jaskaran Singh Saluja is a math wizard with nearly three years of experience as a math teacher. His expertise is in algebra, so he can make algebra classes interesting by turning tricky equations into simple puzzles.
: He loves to play the quiz with kids through algebra to make kids love it.