We can use the following general formula to find the probability of at least three successes in a series of trials:
P(at least 3) = 1 - P(0 successes) - P(1 success) - P(2 successes)
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 three” 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 three.
First, let’s calculate the probability that he makes exactly zero, exactly one, or exactly two free-throws:
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
P(X=2) = 5C2 * .252 * (1-.25)5-2 = 10 * .0625 * .753 = 0.2636
Next, let’s plug these values into the following formula to find the probability that Ty makes at least three free-throws:
- P(X≥3) = 1 – P(X=0) – P(X=1) – P(X=2)
- P(X≥3) = 1 – .2373 – .3955 – .2636
- P(X≥3) = 0.1036
The probability that Ty makes at least three free-throws in five attempts is 0.1036.
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, exactly one, or exactly 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
P(X=2) = 10C2 * .022 * (1-.02)10-2 = 45 * .0004 * .988 = 0.0153
Next, let’s plug these values into the following formula to find the probability that at least three widgets are defective:
- P(X≥3) = 1 – P(X=0) – P(X=1) – P(X=2)
- P(X≥3) = 1 – 0.8171 – 0.1667 – 0.0153
- P(X≥3) = 0.0009
The probability that at least three widgets are defective in this random sample of 10 is 0.0009.
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 three correctly.
First, let’s calculate the probability that he answers exactly zero, exactly one, or exactly two 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
P(X=2) = 5C2 * .602 * (1-.60)5-2 = 10 * .36 * .403 = 0.2304
Next, let’s plug these values into the following formula to find the probability that he answers at least three questions correctly:
- P(X≥3) = 1 – P(X=0) – P(X=1) – P(X=2)
- P(X≥3) = 1 – 0.01024 – 0.0768 – 0.2304
- P(X≥3) = 0.6826
The probability that he answers at least three questions correctly out of five is 0.6826.
Bonus: Probability of “At Least Three” Calculator
Use this calculator to automatically find the probability of “at least three” successes, based on the probability of success in a given trial and the total number of trials.