Geometric Distribution Formula

Geometric Distribution is a probability distribution that tells the number of independent Bernoulli trials needed to achieve the first success. It is used to analyze situations where each trial has only two possible outcomes (success or failure), and the probability of success remains constant across all trials.

What is Geometric Distribution?

The geometric distribution is a probability distribution that describes the number of trials it takes to achieve the first success in a sequence of independent Bernoulli trials, where each trial has only two possible outcomes: success or failure. The distribution is often used in situations where we are interested in the probability of success after a certain number of trials.

Geometric Distribution Formula

The formula for the Geometric Distribution for the number of trials(k) required to get one success:

P (X = k) = (1−p) ^k−1 × p ; k = 1, 2, 3, .......

Where:

• P (X = k) is the probability that the first success occurs on the kth trial,
• p is the probability of success on each trial, and
• k is the number of trials until the first success occurs.

The formula for the Geometric Distribution for the number of failures(k-1) before the first success:

P (X = k) = (1−p) ^k × p ; k = 0, 1, 2, 3, .......

Where:

• P (X = k) is the probability that the first success occurs on the kth trial,
• p is the probability of success on each trial, and
• k is the number of trials before the first success occurs.

Read More,

• Binomial Distribution
• Negative Binomial Distribution
• Probability Theory
• Probability Distribution Function

Solved Examples on Geometric Distribution Formula

Example 1: If a basketball player has a 70% chance of making a free throw. What is the probability that it takes him exactly 3 attempts to make his first free throw?

Solution:

Given:

• Probability of success (making a free throw) p = 0.70
• Number of attempts until the first success k = 3

Using the geometric distribution formula:

P (X = k) = (1−p)^k−1 × p

Substituting the given values:

P (X = 3) = (1−0.70)^{3 − 1} × 0.70

P (X = 3) = (0.30)² × 0.70

P (X = 3) = 0.09 × 0.70

P (X = 3) = 0.063

So, the probability that it takes the basketball player exactly 3 attempts to make his first free throw is 0.063 or 6.3%.

Example 2: Suppose you are flipping a fair coin repeatedly until you get heads. What is the probability that you will need more than 4 flips to get the first heads?

Solution:

In this case, since the coin is fair, the probability of getting heads on each flip is p = 0.5. We want to find the probability of needing more than 4 flips, so we need to find the sum of the probabilities of needing 5 or more flips.

P (X > 4) = 1 − P (X ≤ 4)

P (X > 4) = 1 − [P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)]

P (X > 4) = 1 − [(1−0.5)¹⁻¹ × 0.5 + (1−0.5)²⁻¹ × 0.5 + (1−0.5)³⁻¹ × 0.5 + (1−0.5)⁴⁻¹ × 0.5]

P (X > 4) = 1 − [(0.5) + (0.25) + (0.125) + (0.0625)]

P (X > 4) = 1−0.9375

P (X > 4) = 0.0625

So, the probability that you will need more than 4 flips to get the first heads is 0.0625 or 6.25%.

Example 3: A Software company releases a new version of its app, and, 20% of users upgrade to the new version within the first week. What is the probability that the 5th user to try upgrading will be the first to successfully upgrade?

Solution:

Given:

• Probability of success (user upgrading) p = 0.20
• Number of attempts until the first success k = 5

Using the geometric distribution formula:

P(X = k) = (1−p)^k−1 × p

Substituting the given values:

P(X = 5) = (1−0.20)⁵⁻¹ × 0.20

P(X = 5) = (0.80)⁴ × 0.20

P(X = 5) = 0.4096 × 0.20

P(X = 5) = 0.08192

So, the probability that the 5th user to try upgrading will be the first to successfully upgrade is approximately 0.08192 or 8.192%.

Example 4: A student is taking a multiple-choice quiz with 5 questions. Each question has 4 options, and the student knows the answer to 60% of the questions. What is the probability that the student will need to answer more than 3 questions to get the first question correct?

Solution:

Given:

• Probability of success (student knowing the answer) p = 0.60
• Number of attempts until the first success k > 3

Using the geometric distribution formula:

P(X > 3) = 1 − [P(X = 1) + P(X = 2) + P(X = 3)]

P(X > 3) = 1 − [(1−0.60)¹⁻¹ × 0.60 + (1−0.60)²⁻¹ × 0.60 + (1−0.60)³⁻¹ × 0.60]

P(X > 3) = 1 − [(0.60) + (0.40×0.60) + (0.40 × 0.40 × 0.60)]

P(X > 3) = 1 − (0.60 + 0.24 + 0.096)

P(X > 3) = 1 − 0.936

P(X > 3) = 0.064

So, the probability that the student will need to answer more than 3 questions to get the first question correct is approximately 0.064 or 6.4%.

Practice Questions on Geometric Distribution Formula

Question 1: A fair six-sided die is rolled repeatedly until a 6 is obtained. What is the probability that the first 6 occurs on the third roll?

Question 2: A student is studying for a biology exam. Each time the student takes a practice quiz, there is a 40% chance of getting a question correct. What is the probability that it takes the student at least 5 quizzes to get the first question correct?

Question 3: A factory produces light bulbs, and 8% of the bulbs are defective. What is the probability that the 10th bulb tested is the first defective one found?

Question 4: A basketball player makes free throws with a probability of 75%. What is the probability that it takes the player fewer than 4 attempts to make the first successful free throw?

Question 5: A fisherman catches fish at a certain spot with a probability of 20%. What is the probability that it takes the fisherman exactly 6 tries to catch the first fish?