Probability of choosing health care.

An insurer offers a health plan to the employees of a company. The individual employees may choose exactly two of the supplementary coverages A, B, and C, or they may choose no coverage. The proportions of the company's employees that choose coverages A, B, and C are 1/4, 1/3, and 5/12, respectively. Determine the probability that a randomly chosen employee will choose no supplementary coverage. (Ans 1/2)

I have been attempting this problem for over a week, and I just feel like I keep going in circles...

A = Plan A
B = Plan B
C = Plan C
D = Event that the employee chooses two coverages.

P(A) = 1/4
P(B) = 1/3
P(C) = 5/12

So I have determined that I need to find P(A' B' C').. (A-compliment intersection B-compliment ect..)
However, every time I try to break this down, I just keep going in circles.

Any ideas on how to get going on this problem? Thank you in advance!

I will offer a solution, but you will have to do it in a manner acceptable to your instructor.

Sketch 3 intersecting circles, labeling the circles A,B, C respectfully.
To drop fractions we will assume we have 12 employees, or A =3, (rather than 1/4)
B= 4 employees and C= 5 employees.
(This is done for convenience only, and is not neccessary)

Place a 3 in the top of circle A, a 4 on top of circle B, and a 5 on the bottom of circle C
( this obviously corresponds to the 3/12 , 4/12 and 5/12)

Where the 3 circles intersect place a 0. Why? because no employee may select more than 2 plans, and the intersection is the number of employees that chose all 3 plans.

Let us assume that 1 employee chose plans AB and not C
then place a 1 the intersection of circles AB but outside of the section with a 0 in it.
Since circle A must have 3 in it , then the intersection of circles A and C but outside of the section with a 0 , place a 2.
Then the intersection of circles BC but outside of the section with a 0 must have a 3 .

WE have 3 employees in A , 1 in AB and 2 in AC
We have 4 employees in B, 1 in AB and 3 in BC
We have 5 employees in C 2 in AC and 3 in BC

But the total employees within the circles are only 6,[AB=1 AC=2 BC=3 ]
of our 12 employees 6 chose no plan, 6 chose 2 plans
=========================================================================
If you try other combinations they will not work.
Let AB =2 the AC must be 1 and BC =4 , but this makes circle B have 5, not the mandatory 4
let AB=3 , then AC=0 and BC must = 5 , not the mandatory 4
AB CAN NOT BE GREATER THAN 3

Arthur

Yes, thank you for the reply!

After thinking about it more logically, I was able to come up with an operation that my instructor will accept.

P(A) = 1/4, P(B) = 1/3, P(C) = 5/12

Since people can chose ONLY two plans, or none, all of the sets have zero elements except for (S - AuBuC) and the three intersections (not including the intersection of the three). So I assigned unknowns to each intersection. Example, B contains the two intersections of BnA and BnC, so it includes two unknowns. Those two unknowns add up to the 1/3, which is the P(B) initially given.

x + y = 1/4
y + z = 1/3
x + z = 5/12

2(x + y + z) = 1
x + y + z = 1/2 (this includes every possibility that the employees chose two plans)

Since the problem asks for the probability that an employee chooses no plan, and the probability that they choose two is 1/2, then the probability of not choosing any plan is 1 - 1/2 = 1/2.

I was just so hung up on using set operations that I neglected to look at it in a more logical manner.

Thank you!

I am glad you used Probability rather than set theory.
I am afraid I was prejudiced because you mentioned intersecting circles in the problem, if I remember correctly.
I approached the problem incorrectly but your answer of 1/2 is correct of course.

In mathematics ther is " more than one way to skin a cat "

Best of luck in your math
Arthur