# If I toss a coin 3 times and want to know the probability of at least one head, I have understood th

If I toss a coin 3 times and want to know the probability of at least one head, I have understood that the answer is $1-{0.5}^{3}=99\mathrm{%}$. However, why cannot I not use the additon rule $P\left(A\cup B\right)=P\left(A\right)+P\left(B\right)-P\left(A\cap B\right)$, i.e. $0.5+0.5+0.5-{0.5}^{3}$?
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Angelo Murray
I assume you are talking about the Inclusion-exclusion principle when you say addition rule.
You cannot used the addition rule for this problem because you are discussing 3 coins in your problem. If you were discussing two, the addition rule above would be enough.
The correct formula for three coins would be $P\left(A\cup B\cup C\right)=P\left(A\right)+P\left(B\right)+P\left(C\right)-P\left(A\cap B\right)-P\left(B\cap C\right)-P\left(C\cap A\right)+P\left(A\cap B\cap C\right)$The answers is, therefore $0.5+0.5+0.5-{0.5}^{2}-{0.5}^{2}-{0.5}^{2}+{0.5}^{3}=0.875=1-{0.5}^{3}$
However, in problems like this note that it is better not to use the addition rule.
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Jaqueline Kirby
Make sure you are careful when you define and count your events. For example, for this problem, we can say $A$ is the event of getting EXACTLY ONE HEAD. Let $B$ be the event of getting EXACTLY TWO HEADS. Let C be the event of getting EXACTLY THREE HEADS. Then the probability you are looking for is $P\left(A\right)+P\left(B\right)+P\left(C\right)$. Notice we don't subtract anything because the events are distinct (no overlaps). $P\left(A\right)=3\left(1/2{\right)}^{3}=3/8$, $P\left(B\right)=3\left(1/2{\right)}^{3}=3/8$, and $P\left(C\right)=\left(1/2{\right)}^{3}=1/8$. The sum is 7/8, which is the same as the other way, which gave $1-\left(1/2{\right)}^{3}=7/8$.