How to Find the Probability of “At Least Three” Successes


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.

Leave a Reply

Your email address will not be published.