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) = _{n}C_{k} * p^{k} * (1-p)^{n-k}**

where:

**n:**number of trials**k:**number of successes**p:**probability of success on a given trialthe number of ways to obtain_{n}C_{k}:*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) **= _{5}C_{0} * .25^{0} * (1-.25)^{5-0} = 1 * 1 * .75^{5} =** 0.2373**

**P(X=1) **= _{5}C_{1} * .25^{1} * (1-.25)^{5-1} = 5 * .25 * .75^{4} =** 0.3955**

**P(X=2) **= _{5}C_{2} * .25^{2} * (1-.25)^{5-2} = 10 * .0625 * .75^{3} =** 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) **= _{10}C_{0} * .02^{0} * (1-.02)^{10-0} = 1 * 1 * .98^{10} =** 0.8171**

**P(X=1) **= _{10}C_{1} * .02^{1} * (1-.02)^{10-1} = 10 * .02 * .98^{9} =** 0.1667**

**P(X=2) **= _{10}C_{2} * .02^{2} * (1-.02)^{10-2} = 45 * .0004 * .98^{8} =** 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) **= _{5}C_{0} * .60^{0} * (1-.60)^{5-0} = 1 * 1 * .40^{5} =** 0.01024**

**P(X=1) **= _{5}C_{1} * .60^{1} * (1-.60)^{5-1} = 5 * .60 * .40^{4} =** 0.0768**

**P(X=2) **= _{5}C_{2} * .60^{2} * (1-.60)^{5-2} = 10 * .36 * .40^{3} =** 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.