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In a survey of 248 people, 156 are married, 70 are self-employed and 2

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In a survey of 248 people, 156 are married, 70 are self-employed and 2  [#permalink]

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New post Updated on: 24 Jul 2016, 10:05
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In a survey of 248 people, 156 are married, 70 are self-employed and 25% of those who are married are self-employed. If a person is to be randomly selected from those surveyed, what is the probability that the person selected will be self-employed but not married?

A. 1/8
B. 4/31
C. 117/248
D. 1/4
E. 31/117

Originally posted by gsaxena26 on 14 Jun 2010, 06:35.
Last edited by Bunuel on 24 Jul 2016, 10:05, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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Re: In a survey of 248 people, 156 are married, 70 are self-employed and 2  [#permalink]

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New post 14 Jun 2010, 06:54
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I am not sure if my method is correct, but here's how I think you should approach this problem:

Total number of people = 248
Number of married people = 156
Number of self employed people = 70
Number of people who are married AND self employed = 25% of 156 = 39
Number of people who are self employed but not married = People self employed - People self employed/married

Thus, number of people who are self employed but not married = 70-39 = 31

Thus probability = 31/248 = 1/8

Does this match the correct answer specified?

Also, I've attached a diagrammatic view of how this might work.

@Others: Please let me know if my approach is right!
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File comment: Venn Diagram Version of Problem
ProbI.JPG
ProbI.JPG [ 18.3 KiB | Viewed 2249 times ]

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Re: In a survey of 248 people, 156 are married, 70 are self-employed and 2  [#permalink]

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New post 14 Jun 2010, 06:59
I would put the figures into a table:
[tried to post link to my table - but due to the fact that i dont have enough posts/not long enough member of this forum can't post the link - pm me if you need it.

(Table columns: married, not-married, total
table rows: self-employed, not self-employed & total)

Than you can start to plug in the given values (Total surveyed: 248, total married: 156, total self-employed:70)

Now calculate 25% of 156 (share of self-employed & married)

And you end up with 39. It's quite easy to define all other values - but not needed for the anwer:

39/248=1/8
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Re: In a survey of 248 people, 156 are married, 70 are self-employed and 2  [#permalink]

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New post 14 Jun 2010, 07:36
Many thanks. You have got it right.
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Re: In a survey of 248 people, 156 are married, 70 are self-employed and 2  [#permalink]

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New post 27 Nov 2017, 10:56
gsaxena26 wrote:
In a survey of 248 people, 156 are married, 70 are self-employed and 25% of those who are married are self-employed. If a person is to be randomly selected from those surveyed, what is the probability that the person selected will be self-employed but not married?

A. 1/8
B. 4/31
C. 117/248
D. 1/4
E. 31/117

Attachment:
tableprob.png
tableprob.png [ 8.93 KiB | Viewed 644 times ]

A double matrix works well here.
The calculations are simple and few, but the categories are easier to see in a matrix.

Total: 248 people
Married: 156
Self-employed: 70

1) 25 percent of those who are married are self-employed:

\(\frac{1}{4}\) * 156 = 39

2) Self-employed but not married:

(70 - 39) = 31

3) Probability that a randomly selected person from all 248 will be self-employed but not married (31):

\(\frac{31}{248} = \frac{1}{8}\)

Answer A
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Re: In a survey of 248 people, 156 are married, 70 are self-employed and 2  [#permalink]

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New post 27 Nov 2017, 11:56
gsaxena26 wrote:
In a survey of 248 people, 156 are married, 70 are self-employed and 25% of those who are married are self-employed. If a person is to be randomly selected from those surveyed, what is the probability that the person selected will be self-employed but not married?

A. 1/8
B. 4/31
C. 117/248
D. 1/4
E. 31/117



hi
both self employed and married
= 156 * (25/100) = 39

only self employed = 70 - 39 = 31
probability:

only self employed / total
= 31 / 248
= 1/8

thanks
cheers through the kudos button if this helps
:cool:
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Re: In a survey of 248 people, 156 are married, 70 are self-employed and 2 &nbs [#permalink] 27 Nov 2017, 11:56
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