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At a certain medical conference, there are only doctors and nurses. If

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At a certain medical conference, there are only doctors and nurses. If  [#permalink]

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New post 03 Nov 2018, 09:16
1
4
00:00
A
B
C
D
E

Difficulty:

  85% (hard)

Question Stats:

35% (01:45) correct 65% (01:24) wrong based on 51 sessions

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At a certain medical conference, there are only doctors and nurses. If 4 people are to be randomly chosen from the conference attendees, what is the probability of randomly choosing at least n doctor(s)?
1) The probability of randomly selecting 4 nurses is \(\frac{1}{3}\)
2) n = 1

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Re: At a certain medical conference, there are only doctors and nurses. If  [#permalink]

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New post 03 Nov 2018, 09:46
1
At a certain medical conference, there are only doctors and nurses. If 4 people are to be randomly chosen from the conference attendees, what is the probability of randomly choosing at least n doctor(s)?
1) The probability of randomly selecting 4 nurses is 1/3
2) n = 1



We don't know the value of n or the total number of D and N.

1.
Probablity is:
NC4/D+NC4=0.33

Since we dont know the value of n. Clearly NS(N=1,2,3)


2.
N=1.
NS.

Combining:
Atleast 1= 1- prob all nurses

1- NC4/D+NC4
Hence,
1-0.33= 66.67

Hence combing C ans
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Re: At a certain medical conference, there are only doctors and nurses. If  [#permalink]

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New post 03 Nov 2018, 09:49
At a certain medical conference, there are only doctors and nurses. If 4 people are to be randomly chosen from the conference attendees, what is the probability of randomly choosing at least n doctor(s)?
1) The probability of randomly selecting 4 nurses is 1/3
2) n = 1

Option A: 4 nurses prob = 1/3, so rest is (combination of nurse + doctor or only doctor) = 2/3 . Moreover 1/3 does say that there are total 12 people comprising of nurse and doctor ....we also do not know how many doctors are there in 12 (may be 1, or 2 or 8 max)

now prob of selecting n doctor so it can be 1 doctor , 2 , 3 or 4.... accordingly prob will be different

so A is not sufficient

Option B: n = 1....so we need only 1 doctor but number of people , doctor or nurse we do not know..population size is also not known

Option B is not sufficient

Lets combine A and B, still we can't say because 1 out of how many doctors we do not know as we do not proportion of nurse and doctor in team of 12...may be 4, 8 or 6,6, 3, 9

So E is the answer
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At a certain medical conference, there are only doctors and nurses. If  [#permalink]

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New post 15 Nov 2018, 00:41
Q: At a certain medical conference, there are only doctors and nurses. If 4 people are to be randomly chosen from the conference attendees, what is the probability of randomly choosing at least n doctor(s)?
1) The probability of randomly selecting 4 nurses is 13
1
3

2) n = 1

Solution:
A. Looking at 1.
p(all nurses) = 1/3
p(at least n doc) = cannot be determined from the given info since 'n' can be 0, 1, 2, 3, 4 => INSUFFICIENT.

B. Looking at 2.
n = 1. It doesn't provide any info on the number of doctors/nurses => Insufficient

C. Combining 1 & 2.
=> p(at least 1 doc) = 1- p(no docs)
=> p(at least 1 doc) = 1- p(all nurses)
=> p(at least n doc) = 1- 1/3 = 2/3 = 0.667 SUFFICIENT

Correct Answer is C
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At a certain medical conference, there are only doctors and nurses. If &nbs [#permalink] 15 Nov 2018, 00:41
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