Three fair quarters are tossed, and a tail appears on at least one of them. What is the probability that at least one head appears? Express your answer as a common fraction.

DeluluSunshine Aug 4, 2024

#1**+1 **

First, let's find the total number of cases we work through. We can actually just list them all out.

We have

\((HHT,HTH,HTT, THH, THT, TTH, TTT)\)

The only possibility where a head doesn't appear is TTT. That means that

\((HHT,HTH,HTT, THH, THT, TTH)\) all work.

Since probability is essentially \(\text{Probability} = \frac{\text{# of Favorable Outcomes}}{\text{Total Outcomes}}\)

So we just have \(\frac{6}{7}\)

So 6/7 is our answer.

Thanks! :)

NotThatSmart Aug 4, 2024