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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.

 Aug 4, 2024
 #1
avatar+1918 
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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! :)

 Aug 4, 2024
edited by NotThatSmart  Aug 4, 2024

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