Table of Contents
The probability of “at least two” successes is calculated by subtracting the probability of zero successes from the probability of two successes. This probability is the sum of the probability of exactly two successes and the probability of exactly three successes, and is the sum of all probabilities of two or more successes.
We can use the following general formula to find the probability of at least two successes in a series of trials:
P(at least two successes) = 1 - P(zero successes) - P(one success)
In the formula above, we can calculate each probability by using the following formula for the :
P(X=k) = nCk * pk * (1-p)n-k
where:
- n: number of trials
- k: number of successes
- p: probability of success on a given trial
- nCk: the number of ways to obtain k successes in n trials
The following examples show how to use this formula to find the probability of “at least two” successes in different scenarios.
Example 1: Free-Throw Attempts
Ty makes 25% of his free-throw attempts. If he attempts 5 free-throws, find the probability that he makes at least two.
First, let’s calculate the probability that he makes exactly zero free throws or exactly one free throw:
P(X=0) = 5C0 * .250 * (1-.25)5-0 = 1 * 1 * .755 = 0.2373
P(X=1) = 5C1 * .251 * (1-.25)5-1 = 5 * .25 * .754 = 0.3955
Next, let’s plug these values into the following formula to find the probability that Ty makes at least two free-throws:
- P(X≥2) = 1 – P(X=0) – P(X=1)
- P(X≥2) = 1 – 0.2372 – 0.3955
- P(X≥2) = 0.3673
The probability that Ty makes at least two free-throw in five attempts is 0.3673.
Example 2: Widgets
At a given factory, 2% of all widgets are defective. In a random sample of 10 widgets, find the probability that at least two are defective.
P(X=0) = 10C0 * .020 * (1-.02)10-0 = 1 * 1 * .9810 = 0.8171
P(X=1) = 10C1 * .021 * (1-.02)10-1 = 10 * .02 * .989 = 0.1667
Next, let’s plug these values into the following formula to find the probability that at least two widgets are defective:
- P(X≥2) = 1 – P(X=0) – P(X=1)
- P(X≥2) = 1 – 0.8171 – 0.1667
- P(X≥2) = 0.0162
The probability that at least two widgets are defective in this random sample of 10 is 0.0162.
Example 3: Trivia Questions
Bob answers 60% of trivia questions correctly. If we ask him 5 trivia questions, find the probability that he answers at least two correctly.
First, let’s calculate the probability that he answers exactly zero or exactly one correctly:
P(X=0) = 5C0 * .600 * (1-.60)5-0 = 1 * 1 * .405 = 0.01024
P(X=1) = 5C1 * .601 * (1-.60)5-1 = 5 * .60 * .404 = 0.0768
Next, let’s plug these values into the following formula to find the probability that he answers at least two questions correctly:
- P(X≥2) = 1 – P(X=0) – P(X=1)
- P(X≥2) = 1 – 0.01024 – 0.0768
- P(X≥2) = 0.91296
The probability that he answers at least two questions correctly out of five is 0.91296.
Bonus: Probability of “At Least Two” Calculator
Use to automatically find the probability of “at least two” successes, based on the probability of success in a given trial and the total number of trials.