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

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02 Sep 2013, 02:14

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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?

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

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02 Sep 2013, 10:38

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Expert's post

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

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

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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

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

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07 Sep 2013, 05:47

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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

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

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08 Sep 2013, 02:14

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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. _________________

“Confidence comes not from always being right but from not fearing to be wrong.”

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

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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)?

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

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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. _________________

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

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21 Nov 2013, 07:47

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Expert's post

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).

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

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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)

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

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