an insurer offers a health plan to the employees of a large company. as part of this plan, the individual…

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 companys employees that choose coverages a, b, and c are $\frac{9}{40}$, $\frac{4}{15}$, and $\frac{7}{24}$, respectively. determine the probability that a randomly chosen employee will choose no supplementary coverage. (hint: draw a venn diagram with three sets, and let $x = p(acap b)$. use the fact that 4 of the 8 regions in the venn diagram have a probability of zero.)\ndetermine the probability that a randomly chosen employee will choose no supplementary coverage. choose the correct answer below\na. $\frac{73}{120}$\nb. $\frac{67}{144}$\nc. $\frac{77}{144}$\nd. $\frac{2}{3}$\ne. 0

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 companys employees that choose coverages a, b, and c are $\frac{9}{40}$, $\frac{4}{15}$, and $\frac{7}{24}$, respectively. determine the probability that a randomly chosen employee will choose no supplementary coverage. (hint: draw a venn diagram with three sets, and let $x = p(acap b)$. use the fact that 4 of the 8 regions in the venn diagram have a probability of zero.)\ndetermine the probability that a randomly chosen employee will choose no supplementary coverage. choose the correct answer below\na. $\frac{73}{120}$\nb. $\frac{67}{144}$\nc. $\frac{77}{144}$\nd. $\frac{2}{3}$\ne. 0

Answer

Explanation:

Step1: Rewrite the given probabilities with a common denominator

The probabilities of choosing A, B, and C are $P(A)=\frac{9}{40}$, $P(B)=\frac{4}{15}$, and $P(C)=\frac{7}{24}$. The least - common multiple of 40, 15, and 24 is 120. $P(A)=\frac{9\times3}{40\times3}=\frac{27}{120}$, $P(B)=\frac{4\times8}{15\times8}=\frac{32}{120}$, $P(C)=\frac{7\times5}{24\times5}=\frac{35}{120}$. Since employees can choose exactly two of the coverages or no coverage, and using the principle of inclusion - exclusion for the case of choosing two - set intersections. Let $x = P(A\cap B)$, $y = P(A\cap C)$, $z = P(B\cap C)$. The sum of the probabilities of choosing exactly two coverages is $P(A\cap B)+P(A\cap C)+P(B\cap C)$. We know that $P(A)=x + y$, $P(B)=x+z$, $P(C)=y + z$. Adding $P(A)+P(B)+P(C)$ gives $2(x + y+z)$. $P(A)+P(B)+P(C)=\frac{27 + 32+35}{120}=\frac{94}{120}$. Then $x + y+z=\frac{94}{240}=\frac{47}{120}$.

Step2: Calculate the probability of no coverage

The sum of the probability of choosing exactly two coverages and the probability of choosing no coverage is 1. Let $P(\text{no coverage})=p$. We have $p=1-(x + y+z)$. $p = 1-\frac{47}{120}=\frac{120 - 47}{120}=\frac{73}{120}$.

Answer:

A. $\frac{73}{120}$