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# There are 8 students. 4 of them are men and 4 of them are women. If 4

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Manager
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There are 8 students. 4 of them are men and 4 of them are women. If 4 [#permalink]  06 Apr 2010, 07:30
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55% (hard)

Question Stats:

61% (02:39) correct 39% (02:01) wrong based on 38 sessions
There are 8 students. 4 of them are men and 4 of them are women. If 4 students are selected from the 8 students. What is the probability that the number of men is equal to that of women?

A. 18/35
B. 16/35
C. 14/35
D. 13/35
E. 12/35
[Reveal] Spoiler: OA

Last edited by Bunuel on 17 Jun 2015, 00:53, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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Re: There are 8 students. 4 of them are men and 4 of them are women. If 4 [#permalink]  06 Apr 2010, 07:43
utin wrote:
There are 8 students. 4 of them are men and 4 of them are women. If 4 students are selected from the 8 students. What is the probability that the number of men is equal to that of women?

A.18/35 B16/35 C.14/35 D.13/35 E.12/35

According to me answer shall be= 4/8 * 3/7 * 4/6 * 3/5

probability that the number of men is equal to that of women so 2 men and 2 women =
(4C2*4C2)/8C4 = 18/35 hence A.
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Re: There are 8 students. 4 of them are men and 4 of them are women. If 4 [#permalink]  06 Apr 2010, 08:05
men= women means 2 men 2 women

total such combinations = 4c2 * 4c2 = 4!/2!.2! * 4!/2!.2! = 6*6

total combinations = 8c4 = 8*7*6*5/4*3*2*1 = 70

so probability = 36/70 = 18/35

hence A
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Re: There are 8 students. 4 of them are men and 4 of them are women. If 4 [#permalink]  06 Apr 2010, 15:22
1
KUDOS
utin wrote:
There are 8 students. 4 of them are men and 4 of them are women. If 4 students are selected from the 8 students. What is the probability that the number of men is equal to that of women?

A.18/35 B16/35 C.14/35 D.13/35 E.12/35

According to me answer shall be= 4/8 * 3/7 * 4/6 * 3/5

You calculated probability of one way in which in which 2 men and 2 women can be selected. But there are 6 different ways in which the selection can happen MMWW, MWMW ,MWWM, WWMM, WMWM,WMMW or 4!/2!2! . So you have to multiply the probability with 6.

6 *4/8 * 3/7 * 4/6 * 3/5 = 18/35
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Manager
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Re: There are 8 students. 4 of them are men and 4 of them are women. If 4 [#permalink]  07 Apr 2010, 08:24
crack700 wrote:
utin wrote:
There are 8 students. 4 of them are men and 4 of them are women. If 4 students are selected from the 8 students. What is the probability that the number of men is equal to that of women?

A.18/35 B16/35 C.14/35 D.13/35 E.12/35

According to me answer shall be= 4/8 * 3/7 * 4/6 * 3/5

You calculated probability of one way in which in which 2 men and 2 women can be selected. But there are 6 different ways in which the selection can happen MMWW, MWMW ,MWWM, WWMM, WMWM,WMMW or 4!/2!2! . So you have to multiply the probability with 6.

6 *4/8 * 3/7 * 4/6 * 3/5 = 18/35

Thanks for clearing my confusion!!! ...thanks a lot...
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Re: There are 8 students. 4 of them are men and 4 of them are women. If 4 [#permalink]  09 Apr 2010, 22:43
the first one that I got right today
A i.e. 18/35. I did same as what gurpreet did.
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Re: There are 8 students. 4 of them are men and 4 of them are women. If 4 [#permalink]  04 Feb 2011, 06:21

let's look at the probability of MMWW.

$$\frac{4}{8}*\frac{3}{7}*\frac{4}{6}*\frac{3}{5} = \frac{3}{35}$$

2 men and 2 women can be arranged in $$^4C2 = 6$$ ways.

1. MMWW
2. MWMW
3. MWWM
4. WWMM
5. WMWM
6. WMMW

so total probability $$= \frac{6*3}{35} = \frac{18}{35}$$

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Re: There are 8 students. 4 of them are men and 4 of them are women. If 4 [#permalink]  09 Jul 2011, 20:23
Probability that the number of men is
equal to that of women = 2 men and 2 women are selected

= 4C2 * 4C2/8C4

= 4!/(2!2!) * 4!/(2!2!)/8!/(4!4!)

= (4*3*2!/(2!2!))^2/(8*7*6 *5*4!/4!/4!)

= 36/(2*35)

= 18/35

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Re: There are 8 students. 4 of them are men and 4 of them are women. If 4 [#permalink]  23 Dec 2011, 11:52
The only possibility is that 2 men and 2 women make the group.

$$P=\frac{C^4_2*C^4_2}{C^8_4}=\frac{18}{35}$$

A
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Re: There are 8 students. 4 of them are men and 4 of them are women. If 4 [#permalink]  26 Dec 2011, 05:01
1
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4C2*4C2/8C4 =18/35
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Re: There are 8 students. 4 of them are men and 4 of them are women. If 4 [#permalink]  16 Jun 2015, 17:02
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Re: There are 8 students. 4 of them are men and 4 of them are women. If 4 [#permalink]  16 Jun 2015, 21:41
1
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Expert's post
utin wrote:
There are 8 students. 4 of them are men and 4 of them are women. If 4 students are selected from the 8 students. What is the probability that the number of men is equal to that of women?

A. 18/35
B. 16/35
C. 14/35
D. 13/35
E. 12/35

According to me answer shall be= 4/8 * 3/7 * 4/6 * 3/5

METHOD-1:

Favorable outcomes (i.e. No. of Men= No. of Women =2) = 4C2 * 4C2 = 6*6 = 36

Total ways of selecting 4 out of 8 Students = 8C4 = 8! / (4! * 4!) = 70

Probability = 36/70 = 18/35

[Reveal] Spoiler:
A

METHOD-2: Also see the mistake done by the person who posted this question

Probability of First selected person being Man = 4/8
Probability of Second selected person being Man = 3/7
Probability of First selected person being Woman = 4/6
Probability of Second selected person being Woman = 3/5

i.e. Probability = (4/8)*(3/7)*(4/6)*(3/5) * [4!/(2!*2!)] = 18/35

The important part for the readers is to understand the reason of multiplying [4!/(2!*2!)] here

When we take the probability of each case like we have have taken in this method then it always include ARRANGEMENTS as well and so we have to take every arrangement of the events as well

The arrangement of these 4 events can be done in 4! ways but since the second man can't be selected before 1st so we have to exclude their arrangement by dividing by 2! and similarly since the second Woman can't be selected before 1st so we have to exclude their arrangement by dividing by 2!

[Reveal] Spoiler:
A

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Re: There are 8 students. 4 of them are men and 4 of them are women. If 4   [#permalink] 16 Jun 2015, 21:41
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