Use addition rules for determining probability of at least one head 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 $1-0.5^3=99\%$. However, why cannot I not use the additon rule $P(A\cup B) = P(A) + P(B) - P(A\cap B)$, i.e. $0.5+0.5+0.5-0.5^3$?
 A: 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(A \cup  B \cup C)= P(A)+P(B)+P(C)-P(A \cap B)-P(B \cap C)-P(C \cap A)+P(A \cap B\cap C) $
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.  
More information on the Inclusion Exclusion principle can be seen here
A: 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 $P(A) + P(B) + P(C)$. Notice we don't subtract anything because the events are distinct (no overlaps). $P(A) = 3(1 / 2)^3 = 3/8$, $P(B) = 3(1 / 2)^3 = 3/8$, and $P(C) = (1 / 2)^3 = 1 / 8$. The sum is $7 / 8$, which is the same as the other way, which gave $1 - (1 / 2)^3 = 7 / 8$.
A: First of all, $1-0.5^3=99\%$ is not correct. The correct computation is  $1-0.5^3=87.5\%$.
To answer your question, you can use the addition rule, but you have to account for all the cases.  If you toss a coin three times, then there are 8 possible outcomes, all of which are mutually exclusive, that is, they cannot happen together.
The 8 outcomes are HHH, HHT, HTH, THH, TTH, THT, HTT, TTT.
Of these, the first seven all contain at least 1 head.  So we could say that the probability of at least one head is 
P(HHH) + P(HHT) + P(HTH) + P(THH) + P(TTH) + P(THT) + P(HTT).
Each of these events is connected by "or."
If the coin is fair, then H or T is equally likely, so each individual toss has probability 0.5.  So P(HHH) = $0.5^3$ = 0.125, for example.  The other possibilities are also 0.125, so we get an overall probablility of (7)(0.125) which is 0.875, or 87.5%, which agrees with the first calculation using the complement.
