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A committee of 2 people is to be selected out of 3 teachers and 4 prea

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A committee of 2 people is to be selected out of 3 teachers and 4 prea  [#permalink]

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New post 03 Oct 2016, 23:24
00:00
A
B
C
D
E

Difficulty:

  15% (low)

Question Stats:

81% (01:39) correct 19% (02:01) wrong based on 114 sessions

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Re: A committee of 2 people is to be selected out of 3 teachers and 4 prea  [#permalink]

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New post 04 Oct 2016, 03:29
1
Bunuel wrote:
A committee of 2 people is to be selected out of 3 teachers and 4 preachers. If the committee is selected at random, what is the probability that the committee will be composed of at least 1 preacher?

A. 1/4
B. 1/3
C. 2/3
D. 6/7
E. 8/9



Answer D
Total no of ways in which you can select a committee of 2 people out of 7 people is 7C2 ways= 21 ways
Requirement is at-least 1 preacher= 4C1*3C1+4C2 or 7C2-3C2=18 ways
Probability= 18/21 = 6/7
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Re: A committee of 2 people is to be selected out of 3 teachers and 4 prea  [#permalink]

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New post 29 Nov 2017, 06:15
Can any one explain better why is the D the result ? i just thought it would be prob=4C1*4C2/7C2

Thx in advance
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Re: A committee of 2 people is to be selected out of 3 teachers and 4 prea  [#permalink]

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New post 01 Dec 2017, 07:45
1
2
Bunuel wrote:
A committee of 2 people is to be selected out of 3 teachers and 4 preachers. If the committee is selected at random, what is the probability that the committee will be composed of at least 1 preacher?

A. 1/4
B. 1/3
C. 2/3
D. 6/7
E. 8/9



To determine how many ways to select at least one preacher, we can use the formula:

Number of ways to select at least 1 preacher = total number of ways to select the group - number of ways to select NO preachers

Total number of ways to select the group is:

7C2 = 7!/[2!(7-2)!] = (7 x 6)/2! = 21

The number of ways to select NO preachers is:

3C2 = (3 x 2)/2! = 3

Thus, the number of ways to select at least 1 preacher is 21 - 3 = 18 ways.

So the probability that the committee will be composed of at least 1 preacher is 18/21 = 6/7.

Answer: D
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A committee of 2 people is to be selected out of 3 teachers and 4 prea  [#permalink]

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New post 07 Aug 2019, 02:59
Bit late here, but think my solution will add up to the above ;Incase,you still don’t understand

P(at least 1 preacher) + P(no preacher) =1
.: P(at least 1preacher) =
1 - P(no preacher)

So P(no preacher ) = 3C2/7C2 =1/7

It’s 3C2 because we have 3 teachers who are not preachers of which we select 2
Divided by 7people of which we select 2 ...Got it :)

Now, P(at least 1 preacher) = 1-1/7
P(at least 1 preacher) = 6/7

Answer D :)

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Re: A committee of 2 people is to be selected out of 3 teachers and 4 prea  [#permalink]

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New post 09 Oct 2019, 04:56
Staphyk wrote:
Bit late here, but think my solution will add up to the above ;Incase,you still don’t understand

P(at least 1 preacher) + P(no preacher) =1
.: P(at least 1preacher) =
1 - P(no preacher)

So P(no preacher ) = 3C2/7C2 =1/7

It’s 3C2 because we have 3 teachers who are not preachers of which we select 2
Divided by 7people of which we select 2 ...Got it :)

Now, P(at least 1 preacher) = 1-1/7
P(at least 1 preacher) = 6/7

Answer D :)



7 ! /(7–2) ! 2 ! or 7 ! / (5 ! ) ( 2 ! ) =

5040 / (120)(2) =

7•6•5•4• 3•2•1 / ( 5•4•3•2•1 ) ( 2 • 1 )

= 7 • 6 / 2 • 1 = 42/2 = 21
plz revise this part (So P(no preacher ) = 3C2/7C2 =1/7
)
and as easy approach :

3c2/7c2=3/21=1/7>>>>> at least 1 preacher>>>1-1/7=6/7


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Re: A committee of 2 people is to be selected out of 3 teachers and 4 prea  [#permalink]

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New post 12 Oct 2019, 04:21
Staphyk wrote:
Bit late here, but think my solution will add up to the above ;Incase,you still don’t understand

P(at least 1 preacher) + P(no preacher) =1
.: P(at least 1preacher) =
1 - P(no preacher)

So P(no preacher ) = 3C2/7C2 =1/7

It’s 3C2 because we have 3 teachers who are not preachers of which we select 2
Divided by 7people of which we select 2 ...Got it :)

Now, P(at least 1 preacher) = 1-1/7
P(at least 1 preacher) = 6/7

Answer D :)

Posted from my mobile device


Hello, thanks for the explanation

Do you mind explaining why we are involving combinatorics in probability.

Why didn't we just use the normal probability method i.e. 1 - 3/7

Thanks
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Re: A committee of 2 people is to be selected out of 3 teachers and 4 prea   [#permalink] 12 Oct 2019, 04:21
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