**Probability** tells us the likelihood that some event occurs.

For example, suppose 4% of all students at a certain school prefer math as their favorite subject. If we randomly select one student, the probability that they prefer math would be 4%.

But often we’re interested in probabilities involving *several* trials. For example, if we randomly select three students, what is the probability that **at least one** prefers math?

We can use the following steps to answer this:

**1. Find the probability that a student does not prefer math.**

We know the probability that a student prefers math is P(prefers math) = .04.

Thus, the probability that a student does not prefer math is P(does not prefer math) = .96.

**2. Find the probability that all students selected do not prefer math.**

Since the probability that each student prefers math is independent of each other, we can simply multiply the individual probabilities together:

P(all students do not prefer math) = .96 * .96 * .96 = .8847.

This represents the probability that all three students do not prefer math as their favorite subject.

**3. Find the probability that at least one student prefers math.**

Lastly, the probability that at least one student prefers math is calculated as:

P(at least one prefers math) = 1 – P(all do not prefer math) = 1 – .8847 = **.1153**.

It turns out that we can use the following general formula to find the probability of at least one success in a series of trials:

P(at least one success) = 1 - P(failure in one trial)^{n}

In the formula above, *n* represents the total number of trials.

For example, we could have used this formula to find the probability that at least one student in a random sample of three preferred math as their favorite subject:

P(at least one student prefers math) = 1 – (.96)^{3} = **.1153**.

This matches the answer that we got using the three-step process above.

Use the following examples as additional practice for finding the probability of “at least one” success.

**Related:** How to Find the Probability of “At Least Two” Successes

**Example 1: Free-Throw Attempts**

Mike makes 20% of his free-throw attempts. If he attempts 5 free-throws, find the probability that he makes at least one.

**Solution:**

- P(makes at least one) = 1 – P(misses a given attempt)
^{n} - P(makes at least one) = 1 – (0.80)
^{5} - P(makes at least one) = 0.672

The probability that Mike makes at least one free-throw in five attempts is **0.672**.

**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 one is defective.

**Solution:**

- P(at least one defective) = 1 – P(given widget is not defective)
^{n} - P(at least one defective) = 1 – (0.98)
^{10} - P(at least one defective) = 0.183

The probability that at least one widget is defective in a random sample of 10 is **0.183**.

**Example 3: Trivia Questions**

Bob answers 75% of trivia questions correctly. If we ask him 3 trivia questions, find the probability that he answers at least one incorrectly.

**Solution:**

- P(at least one incorrect) = 1 – P(given answer is correct)
^{n} - P(at least one incorrect) = 1 – (0.75)
^{3} - P(at least one incorrect) = 0.578

The probability that he answers at least one incorrectly is **0.578**.

**Bonus: Probability of “At Least One” Calculator**

Use this calculator to automatically find the probability of “at least one” success, based on the probability of success in a given trial and the total number of trials.

This site is specially awesome

Thanks a lot. Helped much while trying to understand atleast for GMAT.

How did you get .96?

Fantastic explanation! Thank you so much!