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# In a survey of 348 employees, 104 of them are uninsured, 54 work part

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Math Expert
Joined: 02 Sep 2009
Posts: 60515
In a survey of 348 employees, 104 of them are uninsured, 54 work part  [#permalink]

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07 Nov 2019, 01:46
00:00

Difficulty:

75% (hard)

Question Stats:

60% (03:31) correct 40% (03:44) wrong based on 53 sessions

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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) 9/348
(D) 1/8
(E) 41/91

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Re: In a survey of 348 employees, 104 of them are uninsured, 54 work part  [#permalink]

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07 Nov 2019, 09:40
Bunuel 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) 9/348
(D) 1/8
(E) 41/91

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solve usin 2x2 matrix
---- uninsured---insured---total
PT---13----------41------54
FT----91---------203-----294
total---104--------244-----348
probability that the person will neither work part time nor be uninsured = 203/348 = 7/12
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In a survey of 348 employees, 104 of them are uninsured, 54 work part  [#permalink]

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09 Nov 2019, 03:23
1

If you arrive at the 203 / 348 term at the end, do you check the answer choices first or do you immediately cancel the fraction down to it's smallest possible form?

In this case the answer choices were spread so far apart that only two would need a closer consideration, but even (E) can be eliminated as we can see pretty quickly that 203 / 348 > 1/2 whereas 41/91 < 1/2.
Is this how you do it generally or do you prefer to cancel down just to stay sharp and fast for questions in which the answer choices are really close to each other?
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Re: In a survey of 348 employees, 104 of them are uninsured, 54 work part  [#permalink]

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20 Nov 2019, 03:05
Bunuel

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Re: In a survey of 348 employees, 104 of them are uninsured, 54 work part  [#permalink]

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20 Nov 2019, 18:58
1
Bunuel 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) 9/348
(D) 1/8
(E) 41/91

Are You Up For the Challenge: 700 Level Questions

Letting N = the number of individuals who are neither uninsured nor work part time, we can create the equation:

Total = #Uninsured + #Part time - #Both + #Neither

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

190 = -13 + N

203 = N

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

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Re: In a survey of 348 employees, 104 of them are uninsured, 54 work part   [#permalink] 20 Nov 2019, 18:58
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