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# A certain gym has 900 male members and 1200 female members. Three hund

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Math Expert
Joined: 02 Sep 2009
Posts: 49271
A certain gym has 900 male members and 1200 female members. Three hund  [#permalink]

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31 Jan 2017, 10:14
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77% (01:52) correct 23% (02:07) wrong based on 131 sessions

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A certain gym has 900 male members and 1200 female members. Three hundred of the male members are married to women who have memberships at the gym. If one man and then one woman are selected at random, what is the probability that they are a married couple?

A. 0
B. 1/14693
C. 1/3600
D. 1/1050
E. 1

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Senior Manager
Joined: 13 Oct 2016
Posts: 367
GPA: 3.98
A certain gym has 900 male members and 1200 female members. Three hund  [#permalink]

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31 Jan 2017, 10:44
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Bunuel wrote:
A certain gym has 900 male members and 1200 female members. Three hundred of the male members are married to women who have memberships at the gym. If one man and then one woman are selected at random, what is the probability that they are a married couple?

A. 0
B. 1/14693
C. 1/3600
D. 1/1050
E. 1

approach 1 (combinatorics).

$$(\frac{300C1}{900C1}* \frac{1}{1200C1} + \frac{300C1}{1200C1}* \frac{1}{900C1})*\frac{1}{2} = \frac{600}{900*1200} * \frac{1}{2} = \frac{1}{3600}$$

approach 2 (probability).

total # of possible pairs = $$900*1200$$

# of married couples = $$300$$

$$\frac{300}{900*1200} = \frac{1}{3600}$$

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Joined: 01 Apr 2015
Posts: 20
Location: United States
Re: A certain gym has 900 male members and 1200 female members. Three hund  [#permalink]

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27 Feb 2017, 21:00
Hi Vitaliy,

Can you please explain the combinatronics approach? I don't follow the logic.

I do understand the probability approach, however.

Thanks!
-Max
Senior Manager
Joined: 13 Oct 2016
Posts: 367
GPA: 3.98
A certain gym has 900 male members and 1200 female members. Three hund  [#permalink]

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28 Feb 2017, 00:05
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1
mbah191 wrote:
Hi Vitaliy,

Can you please explain the combinatronics approach? I don't follow the logic.

I do understand the probability approach, however.

Thanks!
-Max

Hi Max

I'll start from probability approach.

Number of married couples is 300.

By fundamental counting principle we can generate total number of pairs (man - woman) in 900*1200 ways. We can choose man in 900 ways and woman in 1200 ways.

So our probability will be = number of married pairs/total number of ways to gnerate pairs.

Combinatorics:

There are 300 married couples so there are 300 husbands and 300 wives. We can choose one married man from 300 in 300C1 ways. Total number of ways to choose one man from total 900 is 900C1. When husband was chosen, there is only one way to choose wife from total of 1200C1 ways of choosing woman. But we can start choosing either from men or from women, hence the probability that either of them will be chosen first is 1/2.

Hope this helps.
Regards
Vitaliy
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Posts: 6
Re: A certain gym has 900 male members and 1200 female members. Three hund  [#permalink]

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22 Mar 2017, 16:50
It have to be a combination approach?
I did the chance of a man being married = 300/900 = 1/3
Then the chance of getting his wife = 1/1200 (one specific person among the 1200 group)
As we have 1/3 of getting an married man AND 1/1200 to get his wife so > 1/3 * 1/1200 = 1/3600
Director
Joined: 27 May 2012
Posts: 575
Re: A certain gym has 900 male members and 1200 female members. Three hund  [#permalink]

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06 Sep 2018, 02:48
1
Bunuel wrote:
A certain gym has 900 male members and 1200 female members. Three hundred of the male members are married to women who have memberships at the gym. If one man and then one woman are selected at random, what is the probability that they are a married couple?

A. 0
B. 1/14693
C. 1/3600
D. 1/1050
E. 1

Dear Moderator,
Found this Probability question in the combination section, hope you will do the needful. Thank you.
_________________

- Stne

Math Expert
Joined: 02 Sep 2009
Posts: 49271
Re: A certain gym has 900 male members and 1200 female members. Three hund  [#permalink]

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06 Sep 2018, 02:50
stne wrote:
Bunuel wrote:
A certain gym has 900 male members and 1200 female members. Three hundred of the male members are married to women who have memberships at the gym. If one man and then one woman are selected at random, what is the probability that they are a married couple?

A. 0
B. 1/14693
C. 1/3600
D. 1/1050
E. 1

Dear Moderator,
Found this Probability question in the combination section, hope you will do the needful. Thank you.

_________________
Senior Manager
Joined: 22 Feb 2018
Posts: 321
Re: A certain gym has 900 male members and 1200 female members. Three hund  [#permalink]

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06 Sep 2018, 03:24
Bunuel wrote:
A certain gym has 900 male members and 1200 female members. Three hundred of the male members are married to women who have memberships at the gym. If one man and then one woman are selected at random, what is the probability that they are a married couple?

A. 0
B. 1/14693
C. 1/3600
D. 1/1050
E. 1

OA:C

Probability of Selecting Male (Married couple)$$= \frac{300}{900}$$

Probability of Selecting female Married to male selected earlier$$= \frac{1}{1200}$$

Final probability$$= \frac{300}{900} * \frac{1}{1200} = \frac{1}{3600}$$
_________________

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Manager
Joined: 27 Dec 2016
Posts: 222
Re: A certain gym has 900 male members and 1200 female members. Three hund  [#permalink]

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06 Sep 2018, 18:50
Princ wrote:
Bunuel wrote:
A certain gym has 900 male members and 1200 female members. Three hundred of the male members are married to women who have memberships at the gym. If one man and then one woman are selected at random, what is the probability that they are a married couple?

A. 0
B. 1/14693
C. 1/3600
D. 1/1050
E. 1

OA:C

Probability of Selecting Male (Married couple)$$= \frac{300}{900}$$

Probability of Selecting female Married to male selected earlier$$= \frac{1}{1200}$$

Final probability$$= \frac{300}{900} * \frac{1}{1200} = \frac{1}{3600}$$

Hello,

I was wondering could you please explain how you got 1/1200? 300 men married to 300 women. Shouldn't the probability of selecting a woman is 300/1200? Would greatly appreciate it if you could please help eliminate my doubt!
Intern
Joined: 10 Feb 2017
Posts: 36
Location: Viet Nam
GPA: 3.5
WE: General Management (Education)
Re: A certain gym has 900 male members and 1200 female members. Three hund  [#permalink]

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06 Sep 2018, 23:03
vitaliyGMAT wrote:
Bunuel wrote:
A certain gym has 900 male members and 1200 female members. Three hundred of the male members are married to women who have memberships at the gym. If one man and then one woman are selected at random, what is the probability that they are a married couple?

A. 0
B. 1/14693
C. 1/3600
D. 1/1050
E. 1

approach 1 (combinatorics).

$$(\frac{300C1}{900C1}* \frac{1}{1200C1} + \frac{300C1}{1200C1}* \frac{1}{900C1})*\frac{1}{2} = \frac{600}{900*1200} * \frac{1}{2} = \frac{1}{3600}$$

approach 2 (probability).

total # of possible pairs = $$900*1200$$

# of married couples = $$300$$

$$\frac{300}{900*1200} = \frac{1}{3600}$$

Hi sir, can you explain why we have to * 1/2 at the approach #1. Much appreciated!!!
Senior Manager
Joined: 22 Feb 2018
Posts: 321
A certain gym has 900 male members and 1200 female members. Three hund  [#permalink]

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07 Sep 2018, 01:21
csaluja wrote:
Princ wrote:
Bunuel wrote:
A certain gym has 900 male members and 1200 female members. Three hundred of the male members are married to women who have memberships at the gym. If one man and then one woman are selected at random, what is the probability that they are a married couple?

A. 0
B. 1/14693
C. 1/3600
D. 1/1050
E. 1

OA:C

Probability of Selecting Male (Married couple)$$= \frac{300}{900}$$

Probability of Selecting female Married to male selected earlier$$= \frac{1}{1200}$$

Final probability$$= \frac{300}{900} * \frac{1}{1200} = \frac{1}{3600}$$

Hello,

I was wondering could you please explain how you got 1/1200? 300 men married to 300 women. Shouldn't the probability of selecting a woman is 300/1200? Would greatly appreciate it if you could please help eliminate my doubt!

csaluja
Probability of Selecting Male (Married couple)$$= \frac{300}{900}$$....(1)

We have to select 1 married male out of total 900 males, so we have 300 options.

Probability of Selecting female Married to male selected earlier$$= \frac{1}{1200}$$

There are 300 married women amongst total 1200 female, but as we have to find the probability of a married couple being selected, Number of favorable case for selecting a female who is married to male already selected in (1) is 1. Total number of selecting one female (without any constraint) is 1200.

Another Method of solving

There are total $$300$$ couples.

Favorable case: Selecting $$1$$ couples of $$300$$ couples : $$300C1 = \frac{300!}{299!*1!} = 300$$

Total number of cases : Selecting $$1$$ male out of $$900$$ males and Selecting $$1$$ female out of $$1200$$ females

$$= 900C1 * 1200C1 = \frac{900!}{899!*1!}*\frac{1200!}{1199!*1!}=900*1200$$

Probability of selecting a married couple :$$\frac{Favorable \quad cases}{Total \quad number \quad of \quad cases}= \frac{300}{900*1200} = \frac{1}{3600}$$
_________________

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A certain gym has 900 male members and 1200 female members. Three hund &nbs [#permalink] 07 Sep 2018, 01:21
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