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Of 5 employees, 3 are to be assigned an office and 2 are to [#permalink]
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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
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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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02 Sep 2013, 10:38
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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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
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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
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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08 Sep 2013, 02:14
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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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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21 Nov 2013, 01:45
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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21 Nov 2013, 07:47
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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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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13 Dec 2013, 00:01
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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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!(53)! = (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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