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Re: A certain company holds an annual Labor day lottery by picking one tic [#permalink]
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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]
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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Re: A certain company holds an annual Labor day lottery by picking one tic [#permalink]
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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]
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]
There is an underlying flaw that I have with the problem. Where in the problem does it say that all every worker had a ticket or that all were at the fair? All we know is that there were 50 people at the fair. There should be a clarification something like "of the tickets given out" or "all workers received a ticket."
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Re: A certain company holds an annual Labor day lottery by picking one tic [#permalink]
I have a doubt here: as you can see I have written according to statement 1,and I think it is enough to solve Bunuel

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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 [#permalink]
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