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 binomial distribution:
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.
First, let’s calculate the probability that exactly zero or exactly one 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 this calculator 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.
