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A certain company holds an annual Labor day lottery by picking one tic

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A certain company holds an annual Labor day lottery by picking one tic  [#permalink]

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New post 24 Feb 2015, 11:18
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A
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C
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Difficulty:

  25% (medium)

Question Stats:

72% (01:32) correct 28% (01:08) wrong based on 284 sessions

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A certain company holds an annual Labor day lottery by picking one ticket at random. There are 50 lottery tickets, one for each worker. What is the difference between the chances that the winner will be a married man and the chances that a winner will be a single woman?

(1) There are 30 women in the company and one half of the company workers are married.
(2) There are 10 single men in the company.

Please share the approach applied to arrive at the answer.
Thanks.




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Re: A certain company holds an annual Labor day lottery by picking one tic  [#permalink]

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New post 24 Feb 2015, 22:55
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Hi solitaryreaper,

This question is more about your organizational skills than anything else (the Arithmetic involved is really simple).

We're told that there are 50 lottery tickets. We're asked for the difference in the probability of a married man winning the lottery and a single woman winning the lottery. So we either need to know the exact number of each or something about the ratio of the two (and how it relates to the 50 total people).

Fact 1: There are 30 women and half of the workers are married.

We can do some simple arithmetic based on this information:

Total = 50
Women = 30
Men = 20

Total = 50
Married = 25
Single = 25

Unfortunately, this does not tell us how many married men and single women there are.
Fact 1 is INSUFFICIENT

Fact 2: There are 10 single men in the company

This does not tell us how many single women there are.
Fact 2 is INSUFFICIENT

Combined, we can make a big "table" of the data. Let's start with Married vs. Single:

Married = 25
Single = 25

Since we know that there are 10 single men, we can start to subdivide the categories:

Married = 25
Single = 25
-->Single Men = 10
-->Single Women = 15

We can use this information to subdivide the other categories:

Women = 30
Men = 20
-->Single Men = 10
-->Married Men = 10

Now we know the odds of a married man winning (10/50) and the odds of a single woman winning (15/50), so we can answer the question.
Combined, SUFFICIENT

Final Answer:

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Re: A certain company holds an annual Labor day lottery by picking one tic  [#permalink]

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New post 25 Feb 2015, 01:43
Here we go:

There are 50 lottery tickets, one for each worker.

So there are 50 workers.

St1: There are 30 women in the company and one half of the company workers are married.
25 workers are married.
But we can not deduce anything further from this statement

St2: There are 10 single men in the company.

Clearly not sufficient


Combining:
Married = 25
Unmarried = 25 ----(1)

From statement 2
10 men are single ------> from (1) ---> 15 women are single
There are 30 women in company out of which 15 are single.

So there are 15 women out of 25 married persons, hence 10 men are married.

To Consolidate:

Single Men = 10
Single Women = 15
Married Men = 10
Married Women = 15

Total = 50

So we have all information required to answer the question.

hence option C is correct
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A certain company holds an annual Labor day lottery by picking one tic  [#permalink]

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New post 11 May 2015, 01:13
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Hi guys. Why isn't "A" the answer?

Statement 1 says that 3/5 of company workers are women (3/5= women, 2/5 = men) and says that half of the company workers are married (1/2 = married, 1/2 = not married).

Therefore, the probability of married man winning the lottery is (2/5 * 1/2) = 1/5,
and the probability of single woman winning the lottery is (3/5*1/2) = 3/10.

Hence; the difference in probability is 1/5 - 3/10 = -1/10.

Please enlighten me on which steps I did wrongly. Thank you guys!
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Re: A certain company holds an annual Labor day lottery by picking one tic  [#permalink]

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New post 11 May 2015, 02:35
Xdeatel wrote:
Hi guys. Why isn't "A" the answer?

Statement 1 says that 3/5 of company workers are women (3/5= women, 2/5 = men) and says that half of the company workers are married (1/2 = married, 1/2 = not married).

Therefore, the probability of married man winning the lottery is (2/5 * 1/2) = 1/5,
and the probability of single woman winning the lottery is (3/5*1/2) = 3/10.

Hence; the difference in probability is 1/5 - 3/10 = -1/10.

Please enlighten me on which steps I did wrongly. Thank you guys!


Hi Xdeatel,

Statement-I tells us that combined number of married men and women constitute half of the company workers. It does not tell us that exactly half of the men and half of the women are married. For example: there may be 5 married men and 20 married women equaling 25 married workers. Thus we will have half of the company workers who are married which does not imply that strictly half of the men and half of the women are married.

P(men being married) = P(women being married) = \(\frac{1}{2}\) is based on the assumption that half of the men and half of the women are married which may not be true.

Combining statements- I & II gives us the number of married men and single women. Thus the answer is Option C.

Hope its clear!

Regards
Harsh

Hope its clear!
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Re: A certain company holds an annual Labor day lottery by picking one tic  [#permalink]

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New post 07 Feb 2016, 10:22
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I disagree with a lot of you on this one. The answer is A indeed. See the attached picture.
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File comment: Solution to the problem : "A certain company holds an annual..."
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Re: A certain company holds an annual Labor day lottery by picking one tic  [#permalink]

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New post 11 Apr 2017, 18:53
Help! I need help for such kind of these problems. So, GMAT will not touch on the third sex? I do not offend anybody.
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Re: A certain company holds an annual Labor day lottery by picking one tic  [#permalink]

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Re: A certain company holds an annual Labor day lottery by picking one tic &nbs [#permalink] 04 Oct 2018, 10:50
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