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) = _{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 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) **= _{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**

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) **= _{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**

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) **= _{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**

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.