For questions like:
Dion has an 82.5% chance (0.825 chance) of making a free-throw shot, Dion shoots 4 times, what is the probability of Dion missing at least one free-throw shot?
It is generally accepted that we can calculate P(Dion misses at least one shot) by finding the reciprocal of P(Dion makes all 4 shots), or
P(Misses at least 1) = 1 – P(Makes all shots)
But why is this so? Is there a mathematical proof for this that I can look up to know why this works– why "Makes all shots" and "Misses at least one" are complementary?
Best Answer
Let $A$ be the event 'no misses'. Then $A^c$ is the event 'there is at least one miss'
Recall that a probability measure $\mathbb{P}$ on a space $(\Omega, \mathcal{F})$ satisfies the following two properties:
(1) $\mathbb{P}(\Omega) = 1$
(2) $A , B \in \mathcal{F}, A \cap B = \emptyset \implies \mathbb{P}(A \cup B) = \mathbb{P}(A) + \mathbb{P}(B)$
Thus, using these properties
$$1= \mathbb{P}(\Omega)= \mathbb{P}(A \cup A^c) = \mathbb{P}(A) + \mathbb{P}(A^c)$$ $$\implies \mathbb{P}(\text{at least one miss}) = \mathbb{P}(A^c) = 1- \mathbb{P}(A) = 1- \mathbb{P}(\text{no misses})$$