## Probability of Failure

Let’s revisit the original example:

*You have a single six-sided die. If you roll it once, what is the probability you come up with a 5?*

The probability is 1/6.

Now let’s bring some money into this. You’re in Vegas, and you’re going to win $500 if the die lands on 2. You have a 1/6 or 1 in 6 chance of success.

But what is your chance of failure? Since 5 out of the 6 outcomes would cause failure, you have a 5/6 probability of failure.

Now a different scenario. What if someone were to offer you $500 if, when you roll one die, you rolled **at least a 2? **Think about winning outcomes. If you roll a 2, 3, 4, 5 or 6, you would win. What is the probability of that happening? There are 5 ways to win out of 6, so the **probability of success** would be 5/6.

Here, solving for “at least” or “or” was fairly simple.

probability of rolling at least a 2

= (probability of rolling a 2) + (probability of rolling a 3) + (probability of rolling a 4)

+ (probability of rolling a 5) + (probability of rolling a 6)

= 1/6 + 1/6 + 1/6 + 1/6 + 1/6 = 5/6

By using the phrase “at least,” the problem created more ways to succeed than to fail. Notice that finding this probability required more calculations.

Instead of trying to find the probabilities of the ways to win, you can find the probability of failure and subtract it from 1. For rolling at least a 2, the only way NOT to win is to roll a 1. The probability of rolling a 1 is 1/6. If the probability of failure is 1/6, the probability of success must be 5/6.

As a formula: P(success) + P(failure) = 1 ** or ** P(success) = 1 – P(failure)

## Example

A bag contains 10 red marbles and 6 black marbles. If 3 marbles are pulled from the bag without replacement, what is the probability there will be at least 1 red marble?

(A) 1/28

(B) 5/168

(C) 3/8

(D) 27/28

(E) 167/168