Which distribution has a set number of trials?

Probability and Statistics

The Most Common Discrete Probability Distributions Explained with Examples

Probability Distributions

A probability distribution is a mathematical function that describes the likelihood of obtaining the possible values for an event. A probability distribution may be either discrete or continuous. A discrete distribution is one in which the data can only take on certain values, while a continuous distribution is one in which data can take on any value within a specified range (which may be infinite).

There are a variety of discrete probability distributions. The usage of discrete probability distributions depends on the properties of your data. For example, use the:

• Binomial distribution to calculate probabilities for a process where only one of two possible outcomes may occur on each trial, such as coin tosses.
• Hypergeometric distribution to find the probability of k successes in n draws without replacement.
• Poisson distribution to measure the probability that a given number of events will occur during a given time frame, such as the count of library book checkouts per hour.
• Geometric distribution to determine the probability that a specified number of trials will take place before the first success occurs.

Binomial Distribution

The binomial distribution is probably the most widely known of all discrete distribution. It is a type of distribution that has two possible outcomes. One typical example of using binomial distribution is flipping coins. A coin toss has only two possible outcomes: heads or tails, and each outcome has the same probability of 1/2. Let’s take a look at when the Binomial Distribution can be used!

Main Characteristics of a Binomial Distribution:

• The experiment involves n identical trials.
• Each trial has only two possible outcomes denoted as success or failure.
• The trials are independent of each other.
• Denote p as the probability of success, which remains the same between trails, and q = (1 — p) as the probability of getting a failure on any trial.

Example

Q: A warehouse ships ten printing machines, four of which are defective to a company. This company selects five of the machine at random and accepts the shipment if all five machines are non-defective.

Find the probability that the company accepts the shipments when sampling with replacement.

I’m going to use the formula above:
The number of trials (n) is 5, the probability of getting a defective machine (p) is 4/10, so q is 6/10, and x = 5.

Hypergeometric Distribution

The hypergeometric distribution is a probability distribution that’s very similar to the binomial distribution. Both the hypergeometric distribution and the binomial distribution describe the number of times an event occurs in a fixed number of trials. The probability remains the same for every trial for the binomial distribution. In contrast, in the hypergeometric distribution, each trial changes the probability for each subsequent trial because there is no replacement.

Main Characteristics of a Hypergeometric Distribution:

• Consider a collection of N= N1 + N2 similar objects, N1 of them belong to one of two dichotomous classes, and N2 of them belonging to the second class.
• A collection of n objects selected from these N objects at random and without replacement.

Example

Q: Let’s change the previous problem a little bit. If we now sample without replacement, what’s the probability that the company accepts the shipment?

We know that the total number of printing machines N is 10, and the number of machines chosen at random to be tested n is 5. Let’s define N1 as non-defective, and N2 defective, so N1 =6 and N2= 10- N1 =4. For the company to accept the shipment, we can’t have any defective machine. The number of ways to choose all non-defective machine is 6C5, and the number of ways to select 0 defective machine is 4C0.

Poisson Distribution

Poisson distribution helps us to predict the probability of a specific event occurs within a time interval.

Main Characteristics of a Poisson Distribution:

• The number of changes occurring in nonoverlapping intervals is independent.
• The probability of exactly one change occurring in a sufficiently short interval of length h is approximately λh, where λ>0.
• The probability of two or more changes occurring in a sufficiently short interval is essentially zero.

Notice that the Poisson is a limiting form of the Binomial Distribution. For large n, we have p=λ/n.

Example

Q: Suppose that the probability of suffering a side effect from a certain flu vaccine is 0.005. If 1000 persons are inoculated, find the approximate probability that at most one person suffers.

Since n=1000 is a large number, we can use the Poisson approximate binomial distribution to solve the problem, where λ =pn = 0.005 * 1000 =5.

P(x≤1) = P(x=0)+P(x=1)

Q: Phone calls are received at a certain residence as a Poisson process with parameter λ =2 per hour. If Diane takes a 10-mins shower, what’s the probability that the phone rings during that time?

Given that two calls per 60 mins, we first calculate the number of calls we expect per 10 mins, λ=2 * 10 / 60 = 1/3. Now we want to calculate the probability of getting at least one call during this 10 mins, so essentially we want to compute P(X≥1), which can be written as 1 — P(X=0).

Geometric Distribution

The geometric distribution represents the probability that a specified number of trials will take place before the first success occurs. One typical problem that follows the geometric distribution is to determine the number of times a flipped coin comes up tails before it first comes up heads.

Main Characteristics of a Geometric Distribution:

1. Consider a series of independent trials, each having one of two possible outcomes, success or failure. Let p be the probability of success. Define random variable X to be the trial at which the first success occurs.
2. In theory, the number of trials could go on forever. There must be at least one trial.

Example

Q: The probability that a machine produces a defective item is 0.01. Each item is checked as it is produced. Assume that these are independent trials, and compute the probability that at least 100 items must be checked to find one that is defective.

Using the formula above, P(X≥100) → P(X>99)

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