varitero5w

2022-06-26

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. $0.5+0.5+0.5-{0.5}^{3}$?

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.

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 $1-\left(1/2{\right)}^{3}=7/8$.

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