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# According to study in an office, 12 out of 30 employees were intereste

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Joined: 18 Feb 2019
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GMAT 1: 490 Q47 V13
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According to study in an office, 12 out of 30 employees were intereste  [#permalink]

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05 Mar 2019, 01:35
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According to study in an office, 12 out of 30 employees were interested in doing work from home, if 3 employees were selected random, what is the probability that at most 2 of them were interested in doing work from home?

A. 11/203
B. 192/203
D. 64/65
D. 1/203
E. None of these
Intern
Joined: 26 Dec 2018
Posts: 12
Re: According to study in an office, 12 out of 30 employees were intereste  [#permalink]

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05 Mar 2019, 22:28
I cannot clearly understand how the solution has been arrived at. Someone please explain

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Posts: 53771
Re: According to study in an office, 12 out of 30 employees were intereste  [#permalink]

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05 Mar 2019, 23:21
1
kiran120680 wrote:
According to study in an office, 12 out of 30 employees were interested in doing work from home, if 3 employees were selected random, what is the probability that at most 2 of them were interested in doing work from home?

A. 11/203
B. 192/203
D. 64/65
D. 1/203
E. None of these

We are selecting 3 employees and want at most 2 of them to be interested in doing work from home. At most 2 means 0, 1 or 2. So, any number but 3. Let's calculate the opposite probability and subtract that from 1.

APPROACH 1:

$$P(at \ most \ 2) = 1 - P(3) = 1 - \frac{12}{30}*\frac{11}{29}*\frac{10}{28} = \frac{192}{203}$$.

APPROACH 2:

$$P(at \ most \ 2) = 1 - P(3) = 1 - \frac{12C3}{30C3} = \frac{192}{203}$$.

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Re: According to study in an office, 12 out of 30 employees were intereste  [#permalink]

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06 Mar 2019, 01:36
Bunuel
$$P(at \ most \ 2) = 1 - P(3) why have you subtracted the P of choosing employees who are interested in doing work from home by 1 ; wont that give us P of employees who do not want to work from home? P of employee who want to work from home = 12/30 and P of employee who do not to work from home = 18/30 so the probability that at most 2 of them were interested in doing work from home should be = 18/30 * 12/30 * 11/29 bunuel why is the highlighted part incorrect? Bunuel wrote: kiran120680 wrote: According to study in an office, 12 out of 30 employees were interested in doing work from home, if 3 employees were selected random, what is the probability that at most 2 of them were interested in doing work from home? A. 11/203 B. 192/203 D. 64/65 D. 1/203 E. None of these We are selecting 3 employees and want at most 2 of them to be interested in doing work from home. At most 2 means 0, 1 or 2. So, any number but 3. Let's calculate the opposite probability and subtract that from 1. APPROACH 1: [m]P(at \ most \ 2) = 1 - P(3) = 1 - \frac{12}{30}*\frac{11}{29}*\frac{10}{28} = \frac{192}{203}$$.

APPROACH 2:

$$P(at \ most \ 2) = 1 - P(3) = 1 - \frac{12C3}{30C3} = \frac{192}{203}$$.

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Re: According to study in an office, 12 out of 30 employees were intereste  [#permalink]

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06 Mar 2019, 11:13
The method I did in is much longer but might be helpful
Atmost 2 means selecting 0, 1 , 2.
So case 1: we select 2 employees interested in work from home(wfh) and 1 employee mot interested in wfh
12c2*18c1
Case 2: we select 1 employee interested in work from home(wfh) and 2 employees mot interested in wfh
12c1*18c2
Case 3: we select 0 employees interested in work from home(wfh) and 3 employees mot interested in wfh
12c0*18c3
Therefore probability of atmost 2 people who want wfh being selected in 3 is sum of case 1,2 and 3, divided by total possibility of choosing any 3 people.
(12c2*18c1+12c1*18c2+12c0*18c3)/30c3
= 192/203
IMO B

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Re: According to study in an office, 12 out of 30 employees were intereste  [#permalink]

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07 Mar 2019, 09:41
AjjayKannan wrote:
I cannot clearly understand how the solution has been arrived at. Someone please explain

Posted from my mobile device

question asks if we select 3 out of 30, what is probablity that at most 2 of them were interested in WfH
we can answer this by finding probablity of all of the selected are interested in WfH and subtracting it from 1. --> P(A) = 1 - P(A')

interested =12, not interested = 18
Number of ways to select 3 from 30 = 30C3
Number of ways to select 3 from 12 interested = 12C3

Probablity of all 3 selected are interested = 12C3/30C3

Probablity of at most two of the selected are interested = 1 - 12C3/30C3 = 192/203
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Re: According to study in an office, 12 out of 30 employees were intereste  [#permalink]

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10 Mar 2019, 19:50
kiran120680 wrote:
According to study in an office, 12 out of 30 employees were interested in doing work from home, if 3 employees were selected random, what is the probability that at most 2 of them were interested in doing work from home?

A. 11/203
B. 192/203
D. 64/65
D. 1/203
E. None of these

Note that “at most 2” means “2 or fewer.” We can use the following formula:

P(at most 2 of them were interested in working from home) = 1 - P(all 3 of them were interested in working from home)

We have:

P(all 3 of them were interested in working from home) = 12/30 x 11/29 x 10/28 = 3/3 x 11/29 x 1/7 = 11/203

Therefore,

P(at most 2 of them were interested in working from home) = 1 - 11/203 = 203/203 - 11/203 = 192/203

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Re: According to study in an office, 12 out of 30 employees were intereste   [#permalink] 10 Mar 2019, 19:50
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