Answer:
![0.85 = P(C) + 0.75 -0.75 P(C)](https://img.qammunity.org/2021/formulas/mathematics/college/zkwh9iu085qq1x3foz23wt5ru54z1d3frn.png)
![0.1 = 0.25 P(C)](https://img.qammunity.org/2021/formulas/mathematics/college/g7mgc23vhuvptr03f0u4bdqa5vdty64r2d.png)
![P(C) = 0.4](https://img.qammunity.org/2021/formulas/mathematics/college/7cmjga9fyyjju7hffgx5lfcrnbi7v8jac2.png)
Explanation:
First we can define some notation useful:
C ="represent the event of incurring in operating charges"
R= represent the event of emergency rooms charges"
For this case we are interested on P(C) since they want "the probability that a claim submitted to the insurance company includes operating room charges."
We have some probabilities given:
![P(R') = 0.25 , P(C \cup R) =0.85](https://img.qammunity.org/2021/formulas/mathematics/college/hl8ox93s4v6144smmny1kg9k040mgchh56.png)
Solution to the problem
By the complement rule we have this:
![P(R') = 0.25 =1-P(R)](https://img.qammunity.org/2021/formulas/mathematics/college/z7ollw8a0mv34bmj4cpgyroxoxx53r6xar.png)
![P(R) = 1-0.25 = 0.75](https://img.qammunity.org/2021/formulas/mathematics/college/qodiirfd1aldfb3kuy8o14sgqgjzhr80h9.png)
Since the two events C and R are considered independent we have this:
![P(C \cap R) = P(C) *P(R)](https://img.qammunity.org/2021/formulas/mathematics/college/thhr0zewfpyipm5vvojikscdcckbqagsv7.png)
Now we can use the total probability rule like this:
![P(C \cup R) = P(C) + P(R) - P(R)*P(C)](https://img.qammunity.org/2021/formulas/mathematics/college/k6kfhbuajikdzseqbl8ur4z44akabkwl30.png)
And if we replace we got:
![0.85 = P(C) + 0.75 -0.75 P(C)](https://img.qammunity.org/2021/formulas/mathematics/college/zkwh9iu085qq1x3foz23wt5ru54z1d3frn.png)
![0.1 = 0.25 P(C)](https://img.qammunity.org/2021/formulas/mathematics/college/g7mgc23vhuvptr03f0u4bdqa5vdty64r2d.png)
![P(C) = 0.4](https://img.qammunity.org/2021/formulas/mathematics/college/7cmjga9fyyjju7hffgx5lfcrnbi7v8jac2.png)