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Of 5 employees, 3 are to be assigned an office and 2 are to

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Of 5 employees, 3 are to be assigned an office and 2 are to  [#permalink]

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New post Updated on: 02 Sep 2013, 02:16
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Of 5 employees, 3 are to be assigned an office and 2 are to be assigned a cubicle. If 3 of the employees are men and 2 are women, and if those assigned an office are to be chosen at random, what is the probability that the offices will be assigned to 2 of the men and 1 of the women?

(A) 1/3
(B) 2/5
(C) 1/2
(D) 3/5
(E) 2/3

Originally posted by arakban99 on 02 Sep 2013, 02:14.
Last edited by Bunuel on 02 Sep 2013, 02:16, edited 1 time in total.
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Re: Of 5 employees, 3 are to be assigned an office and 2 are to  [#permalink]

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New post 02 Sep 2013, 10:38
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arakban99 wrote:
Of 5 employees, 3 are to be assigned an office and 2 are to be assigned a cubicle. If 3 of the employees are men and 2 are women, and if those assigned an office are to be chosen at random, what is the probability that the offices will be assigned to 2 of the men and 1 of the women?

(A) 1/3
(B) 2/5
(C) 1/2
(D) 3/5
(E) 2/3


Probability = \(\frac{Desired Outcomes}{Total Outcomes}\)

Event = Selection of 3 employees to assign office from 5 such that 2 are men and 1 is woman

Total Outcomes = Selection of any 3 employees from 5 employees = 5 C 3 = 10

Desired Outcomes = Selection of 2 Men from 3 Men AND selection of 1 woman from 2 Women. = (3 C 2) * (2 C 1) = 3*2 = 6

Probability = \(\frac{6}{10}\) --------> \(\frac{3}{5}\)
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Re: Of 5 employees, 3 are to be assigned an office and 2 are to  [#permalink]

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New post 06 Sep 2013, 23:06
i guessed that it at least more than 1/2, eliminated ABC and chose D. Combination approach looks best because i got crazy to do probability one. I used reverse one, which is
1 - (3/5*2/4*1/3+1/5*1/4*1+3/5*1/4*1/3+2/5*3/4*1/3)= 1- (1/10+1/10+1/10+1/10)=6/10=3/5

write if it is right
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Re: Of 5 employees, 3 are to be assigned an office and 2 are to  [#permalink]

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New post 07 Sep 2013, 05:47
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1
Temurkhon wrote:
i guessed that it at least more than 1/2, eliminated ABC and chose D. Combination approach looks best because i got crazy to do probability one. I used reverse one, which is
1 - (3/5*2/4*1/3+1/5*1/4*1+3/5*1/4*1/3+2/5*3/4*1/3)= 1- (1/10+1/10+1/10+1/10)=6/10=3/5

write if it is right


Well, your method is good, but I think this is a more safe method:

The committee should look like this : Man, Man, Woman

how to choose 3 people to be assigned in office from 5: 5C3 = 5!/3!2! = 10

How to choose 2 man and 1 woman= 3C2 * 2C1 = 3!/2! * 2!/1! = 6

Probability = Number of Desired outcome /Total possible outcome = 6/10 = 3/5
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Re: Of 5 employees, 3 are to be assigned an office and 2 are to  [#permalink]

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New post 08 Sep 2013, 02:14
2
offices can be allocated as m,m,w or m,w,m or w,m,m
P(m,m,w) = 3/5*2/4*2/3
P(m,w,m) = 3/5*2/4*2/3
P(w,m,m) = 2/5*3/4*2/4

Since its 'OR', need to add all the 3 probabilities. So answer is 1/5+1/5+1/5 = 3/5
somehow am not comfortable with combinations, hence chose to take a direct method.
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Re: Of 5 employees, 3 are to be assigned an office and 2 are to  [#permalink]

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New post 06 Nov 2013, 16:47
Narenn wrote:
arakban99 wrote:
Of 5 employees, 3 are to be assigned an office and 2 are to be assigned a cubicle. If 3 of the employees are men and 2 are women, and if those assigned an office are to be chosen at random, what is the probability that the offices will be assigned to 2 of the men and 1 of the women?

(A) 1/3
(B) 2/5
(C) 1/2
(D) 3/5
(E) 2/3


Probability = \(\frac{Desired Outcomes}{Total Outcomes}\)

Event = Selection of 3 employees to assign office from 5 such that 2 are men and 1 is woman

Total Outcomes = Selection of any 3 employees from 5 employees = 5 C 3 = 10

Desired Outcomes = Selection of 2 Men from 3 Men AND selection of 1 woman from 2 Women. = (3 C 2) * (2 C 1) = 3*2 = 6

Probability = \(\frac{6}{10}\) --------> \(\frac{3}{5}\)


Hi Narenn, can you explain why the total outcomes is not 5! , (the number of ways to arrange 5 different objects)?
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Re: Of 5 employees, 3 are to be assigned an office and 2 are to  [#permalink]

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New post 21 Nov 2013, 01:45
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oss198 wrote:
Narenn wrote:
arakban99 wrote:
Of 5 employees, 3 are to be assigned an office and 2 are to be assigned a cubicle. If 3 of the employees are men and 2 are women, and if those assigned an office are to be chosen at random, what is the probability that the offices will be assigned to 2 of the men and 1 of the women?

(A) 1/3
(B) 2/5
(C) 1/2
(D) 3/5
(E) 2/3


Probability = \(\frac{Desired Outcomes}{Total Outcomes}\)

Event = Selection of 3 employees to assign office from 5 such that 2 are men and 1 is woman

Total Outcomes = Selection of any 3 employees from 5 employees = 5 C 3 = 10

Desired Outcomes = Selection of 2 Men from 3 Men AND selection of 1 woman from 2 Women. = (3 C 2) * (2 C 1) = 3*2 = 6

Probability = \(\frac{6}{10}\) --------> \(\frac{3}{5}\)


Hi Narenn, can you explain why the total outcomes is not 5! , (the number of ways to arrange 5 different objects)?



the question is asking for the probability of SELECTING 2 men & 1 woman, it does not matter if the selection includes woman A OR B.
hope it helps.
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Re: Of 5 employees, 3 are to be assigned an office and 2 are to  [#permalink]

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New post 21 Nov 2013, 07:47
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oss198 wrote:
Hi Narenn, can you explain why the total outcomes is not 5! , (the number of ways to arrange 5 different objects)?


Hi oss198,
Sorry, I saw your post today only.

Arrangement is out of context here. We are finding the ways of selecting the 3 persons out of 5. (either you choose ABC or BCA from ABCDE is same here)

Selection of 3 persons from 5 :- 5 C 3
selection of 3 persons from 5 and arranging them on 3 chairs :- 5 C 3 * 3!
Selection of 5 persons from 5 :- 5 C 5
Selection of 5 persons from 5 and arranging them on 5 chairs :- 5 C 5 * 5! --------> which ultimately turns out to be 5!

If you need brief insights of these concepts check my article, Permutations and Combinations (In my Signature).

Hope that helps! :)
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Re: Of 5 employees, 3 are to be assigned an office and 2 are to  [#permalink]

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New post 23 Nov 2013, 07:59
Thanks a lot for your answers Aldossari and Narenn.
Narenn your signature is awesome :) i'll have a look.
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Re: Of 5 employees, 3 are to be assigned an office and 2 are to  [#permalink]

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New post 13 Dec 2013, 00:01
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1) 3/5*2/4*2/3=2/10 (where 3/5 probability of 1 man being selected, 2/4 the other man, 2/3 one woman)
2) 2/10*3=3/5 (multiplying by 3 as there are 3 cases in different order)
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Re: Of 5 employees, 3 are to be assigned an office and 2 are to  [#permalink]

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New post 06 Apr 2017, 09:35
arakban99 wrote:
Of 5 employees, 3 are to be assigned an office and 2 are to be assigned a cubicle. If 3 of the employees are men and 2 are women, and if those assigned an office are to be chosen at random, what is the probability that the offices will be assigned to 2 of the men and 1 of the women?

(A) 1/3
(B) 2/5
(C) 1/2
(D) 3/5
(E) 2/3


When 3 of the 5 employees are assigned an office, the remaining 2 employees automatically will be assigned a cubicle. Therefore, the number of ways to choose 3 employees to be assigned an office is 5C3 = 5!/3!(5-3)! = (5 x 4 x 3)/(3 x 2 x 1) = 10.

Now we need to find the number of ways in which 2 of the 3 employees selected to have an office will be male and 1 of them will be female. Since there are 3 male employees and 2 female employees, the number of ways to choose 2 male employees out of 3 males and 1 female employee out of 2 females is 3C2 x 2C1 = 3 x 2 = 6.

Finally, since there are 10 ways to choose 3 employees from 5, and 6 ways to choose 2 male employees from 3 and 1 female employee from 2, the probability is 6/10 = 3/5.

Answer: D
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Re: Of 5 employees, 3 are to be assigned an office and 2 are to  [#permalink]

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Re: Of 5 employees, 3 are to be assigned an office and 2 are to   [#permalink] 10 Jun 2019, 06:07
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