Determine the probability that a randomly chosen employee will choose no supplementary coverage.
An insurer offers a health plan to the employees of a large company. As part of this plan, the individual employees may choose exactly two of the supplementary coverages $A, B$, and $C$, or they may choose no supplementary coverage. The proportions of the company's employees that choose coverages $A, B,$ and $C$ are $\frac{1}{4},\frac{1}{3},$ and $\frac{5}{12}$ respectively.
Attempt:
If $P(A) = 3/12$, $P(B) = 4/12$, $P(C) = 5/12$, $P(A)+P(B)+P(C) = 3/12 + 4/12 + 5/12 = 1.$ Got stuck.
Two question:
the answer should be $1 – P(A \cap B) + P(B \cap C) + P(B \cap C) =$ no coverage but
- Can I derive $P(A \cap B)$ from this information?
- is $A=3/12$ correct?
Best Answer
You are given $\mathsf P(A)=\tfrac 14, \mathsf P(B)=\tfrac 13, \mathsf P(C)=\tfrac 5{12}$, but also know $\mathsf P(A\cap(B\cup C)^\complement)=0=\mathsf P(B\cap(A\cup C)^\complement)=\mathsf P(C\cap(A\cup B)^\complement)$
You also know $\mathsf P(A\cap B\cap C)=0$
I suggest drawing a Venn diagram to see that:
$$\mathsf P(A) = \mathsf P(A\cap B)+\mathsf P(A\cap C) \\ \mathsf P(B) = \mathsf P(A\cap B)+\mathsf P(B\cap C) \\ \mathsf P(C) = \mathsf P(A\cap C)+\mathsf P(B\cap C)$$
As the only regions that are not of zero probability are where exactly two sets overlap, and the area outside all three. The areas covered by only one set, or by all three, have zero probability.
Which means that $\mathsf P(A^\complement\cap B^\complement\cap C^\complement) = 1-\color{purple}{\boxed{\qquad\qquad?}}$