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by gmatblood » Sun Nov 06, 2011 11:12 am
In a survey of 348 employees, 104 of them are uninsured, 54 work part time, and 12.5 percent of employees who are uninsured work part time. If a person is to be randomly selected from those surveyed, what is the probability that the person will neither work part time nor be uninsured?

(A) 7/12
(B) 8/41
(C) 91/348
(D) 1/8
(E) 41/91
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by vaibhavgupta » Sun Nov 06, 2011 11:28 am
gmatblood wrote:In a survey of 348 employees, 104 of them are uninsured, 54 work part time, and 12.5 percent of employees who are uninsured work part time. If a person is to be randomly selected from those surveyed, what is the probability that the person will neither work part time nor be uninsured?

(A) 7/12
(B) 8/41
(C) 91/348
(D) 1/8
(E) 41/91
IMO A

12.5% of 104=13
hence people who are neither= 203
probability that the person will neither work part time nor be uninsured=203/348
If OA is A, IMO B
If OA is B, IMO C
If OA is C, IMO D
If OA is D, IMO E
If OA is E, IMO A

FML!! :/
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by Scott@TargetTestPrep » Tue Nov 12, 2019 7:03 pm
gmatblood wrote:In a survey of 348 employees, 104 of them are uninsured, 54 work part time, and 12.5 percent of employees who are uninsured work part time. If a person is to be randomly selected from those surveyed, what is the probability that the person will neither work part time nor be uninsured?

(A) 7/12
(B) 8/41
(C) 91/348
(D) 1/8
(E) 41/91
We can create the equation:

348 = 104 + 54 - 0.125(104) + N

190 = -13 + N

203 = N

P(neither) = 203/348 = 7/12

Answer: A

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