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# From a Group of 8 People, Including George and Nina, 3 people are

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Re: From a Group of 8 People, Including George and Nina, 3 people are [#permalink]
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mohitmohit11
From a Group of 8 People, Including George and Nina, 3 people are to be selected at random to work on a certain project. What is the probability that 3 people selected will include George but not Nina

A 5/56
B 9/56
C 15/56
D 21/56
E 25/56

Similar question to practice: from-a-group-of-8-volunteers-including-andrew-and-karen-160883.html
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From a Group of 8 People, Including George and Nina, 3 people are [#permalink]
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G _ _ | N _ _ _ _ _

This is the configuration that we want. Let's choose them one at a time.

In the first group, we want George and then 2 other people who are not Nina. That's $$\frac{1}{8}*\frac{6}{7}*\frac{5}{6}$$, with 3 ways to arrange George in that group. Notice that the order/selection of the second group doesn't matter at all since we've already accounted for them in the first group.

In total, that's $$\frac{3*6*5}{8*7*6}$$ = $$\frac{3*5}{8*7}= \frac{15}{56}$$ .

Originally posted by GSBae on 05 May 2016, 10:24.
Last edited by GSBae on 19 Nov 2016, 08:19, edited 1 time in total.
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Re: From a Group of 8 People, Including George and Nina, 3 people are [#permalink]
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Let's consider A,B,C,D,E,F,G,H

Proba = $$\frac{Number of outcomes with A and not B }{Total number of outcomes}$$

Total number of outcomes = 3 from 8 = $$\frac{8 * 7 * 6 }{3 * 2 * 1}$$ = 56

Number of outcomes with A and not B

There is 3 cases :

A_ _ : $$\frac{1 * 6 * 5 }{3 * 2 *1}$$ = 5

_A_ : $$\frac{6 * 1 * 5 }{3 * 2 *1}$$ = 5

_ _A : $$\frac{6 * 5 * 1 }{3 * 2 *1}$$ = 5

So Proba = $$\frac{Number of outcomes with A and not B }{Total number of outcomes}$$

= $$\frac{15 }{56}$$
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From a Group of 8 People, Including George and Nina, 3 people are [#permalink]

I don't understand the Combinations method, it makes no sense to me what so ever.

So I am trying to use Venn diagram here.

So: P(N)+P(G)-P(N and G) = 1

G-George N-nina
We want to know George only, so no Nina and no overlap George plus Nina.

P(N)=1-P(not N) $$1- \frac{7}{8}*\frac{6}{7}*\frac{5}{6} = \frac{3}{8}$$
P(G)=1-P(not G) $$1- \frac{6}{7}*\frac{5}{6}*\frac{4}{5} = \frac{3}{7}$$
P(N and G)= P(A) * P(B) $$\frac{9}{56}$$

So: 1-P(N)+P(N and G) = P(G)

$$1-\frac{56}{56}-\frac{21}{56}+\frac{9}{56}$$ SHOULD BE P(G), but it's $$\frac{26}{56}$$

How come????? What is wrong here?...

Thanks!
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Re: From a Group of 8 People, Including George and Nina, 3 people are [#permalink]
iliavko

I don't understand the Combinations method, it makes no sense to me what so ever.

So I am trying to use Venn diagram here.

So: P(N)+P(G)-P(N and G) = 1

G-George N-nina
We want to know George only, so no Nina and no overlap George plus Nina.

P(N)=1-P(not N) $$1- \frac{7}{8}*\frac{6}{7}*\frac{5}{6} = \frac{3}{8}$$
P(G)=1-P(not G) $$1- \frac{6}{7}*\frac{5}{6}*\frac{4}{5} = \frac{3}{7}$$
P(N and G)= P(A) * P(B) $$\frac{9}{56}$$

So: 1-P(N)+P(N and G) = P(G)

$$1-\frac{56}{56}-\frac{21}{56}+\frac{9}{56}$$ SHOULD BE P(G), but it's $$\frac{26}{56}$$

How come????? What is wrong here?...

Thanks!

Hi iliavko,

Why are you using Venn diagram ?
This tool is rather used for sets questions. OK ?

The question asks the probability of choosing a group of 3 people from 8, including A and excluding B.

So, the basic proba formula is P = (Number of outcomes where the event occurs) / (Number total of outcomes)

Number of outcomes where the event occurs
Stage 1 : Number oy ways to choose A = 1
Stage 2 : Number of ways to choose 2 people form 6 (8, minus A, minus B since you want A and no B) = 15
Number of outcomes where the event occurs = 15

Number total of outcomes
Number of ways choose 3 people from 8 = 56

So P = 15/56

OK ?
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Re: From a Group of 8 People, Including George and Nina, 3 people are [#permalink]
Hi Alex,

Yes but Venn diagram can be used here as far as I understand... What you explained is the combinations method and yes its simple and people use it all the time, but I want to know why my logic with Venn diagram here isn't working.
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Re: From a Group of 8 People, Including George and Nina, 3 people are [#permalink]
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iliavko

I don't understand the Combinations method, it makes no sense to me what so ever.

So I am trying to use Venn diagram here.

So: P(N)+P(G)-P(N and G) = 1

G-George N-nina
We want to know George only, so no Nina and no overlap George plus Nina.

P(N)=1-P(not N) $$1- \frac{7}{8}*\frac{6}{7}*\frac{5}{6} = \frac{3}{8}$$
P(G)=1-P(not G) $$1- \frac{6}{7}*\frac{5}{6}*\frac{4}{5} = \frac{3}{7}$$
P(N and G)= P(A) * P(B) $$\frac{9}{56}$$

So: 1-P(N)+P(N and G) = P(G)

$$1-\frac{56}{56}-\frac{21}{56}+\frac{9}{56}$$ SHOULD BE P(G), but it's $$\frac{26}{56}$$

How come????? What is wrong here?...

Thanks!

Hi,

You are going wrong in execution of your DATA..

P(N)=1-P(not N) $$1- \frac{7}{8}*\frac{6}{7}*\frac{5}{6} = \frac{3}{8}$$
P(G)=1-P(not G) $$1- \frac{6}{7}*\frac{5}{6}*\frac{4}{5} = \frac{3}{7}$$
P(N and G)= P(A) * P(B) $$\frac{9}{56}$$

we are looking for - What is the probability that 3 people selected will include George but not Nina?

so It is P( G but not N) = P(G) - P(N and G), where P(G) is when George is there and P(N and G) is when both are together..
so Using your calculations = $$\frac{3}{7} - \frac{9}{56} = \frac{24-9}{56} = \frac{15}{56}$$
Ofcourse it is better to do with combinations but you have used Venn correctly upto a point and then faultered in final execution..

Hope it helps you and this is what you were looking for..
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Re: From a Group of 8 People, Including George and Nina, 3 people are [#permalink]
Hi Chetan!

Cool!, Thank you so much for the clarification!

So I guess the way I did it I was double-counting something somewhere?.. Anyways, what you said makes sense, just grab the P(G) subtract the overlap p(G and N) and get the "pure" P(G).

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Re: From a Group of 8 People, Including George and Nina, 3 people are [#permalink]
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iliavko
Hi Chetan!

Cool!, Thank you so much for the clarification!

So I guess the way I did it I was double-counting something somewhere?.. Anyways, what you said makes sense, just grab the P(G) subtract the overlap p(G and N) and get the "pure" P(G).

Hi iliavko thanks for pointing out this point of view ... !! I was obviously missing something !

According to Vienn Diagramm :
P(A and not B) = P(A) - P(A and B) = P(A) - P(A)*P(B) = P(A) * [1- P(B)] = P(A) * P(not B)

P(A) = $$\frac{3}{8}$$
P(not B) = $$\frac{5}{8}$$

So, P(A) * P(not B) = $$\frac{5}{8}$$ * $$\frac{3}{8}$$ = $$\frac{15}{64}$$

Can you explain why it's wrong ?

Thank you !!
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Re: From a Group of 8 People, Including George and Nina, 3 people are [#permalink]
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mohitmohit11
From a Group of 8 People, Including George and Nina, 3 people are to be selected at random to work on a certain project. What is the probability that 3 people selected will include George but not Nina

A 5/56
B 9/56
C 15/56
D 21/56
E 25/56

We are given that 3 people are to be selected from 8 people and need to determine the probability that, of the 3 people selected, George is included but Nina is not. Let’s first determine the total number of ways to select 3 people from a group of 8.

The number of ways to select 3 people from a group of 8 is 8C3 = (8 x 7 x 6)/3! = 56.

Next, let’s determine the number of ways to select 3 people from a group of 8 when George is included but Nina is not.

Since George must be included and Nina is not, we can reduce the number of available spots from 3 to 2 (because George is already selected), and we can reduce the number of people available to be selected from 8 to 6 (Nina is not even considered, and George is already selected).

Thus, the number of ways to select 3 people when George is included and Nina is not is:

6C2 = (6 x 5)/2! = 15

So, finally, the probability of selecting a group of three with George and not Nina is:

15/56

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Re: From a Group of 8 People, Including George and Nina, 3 people are [#permalink]
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Dear Friends there are two methods to solve this problem: 1) to count the number of unique 3 people groups (denominator) and the number of 3 people groups that include George but not Nina (numerator) 2) Using probability formula.

Since first method has been clearly explained in previous comments I will focus on the second method

1) 1/8 x 6/7 x 5/6 = 5/56 George selected and Nina is not selected

2) 6/8 x 1/7 x 5/6 = 5/56 First is the probability that neither George nor Nina selected. Why it is 6/8? Because there are 8 candidates, but we want those people who are not George nor Nina. Second is the probability that George is selected. The third is probability that Nina is not selected

3) 6/8 x 5/7 x 1/6 = 5/56 First is the probability that neither George nor Nina will be selected. Second is probability that neither George nor Nina will be selected. Why 5/7? Again because after selecting one person for the first slot, there are 7 people remaining (including George and Nina) We want only 5 of them who are neither George nor Nina

5/56 + 5/56 + 5/56 = 15/56
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Re: From a Group of 8 People, Including George and Nina, 3 people are [#permalink]
How come we do not have to multiply the answer by 2:

Total ways of picking 3 people from 8 = 8C3 = 56

Pick George first x Other 2 people excluding Nina = 1 x 6C2 = 15

We can also pick the other 2 people first and George last = 1 x 6C2 = 15

Wouldn't that make the answer 30/56?

When to multiply and when not to is where I'm having trouble. I'm assuming it's because the order doesn't matter in this question. Whether it's GXX or XXG etc. counts as 1 way
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Re: From a Group of 8 People, Including George and Nina, 3 people are [#permalink]
Can someone please explain where I'm going wrong?
If we only have to select George and not Nina out of 8 people that means we need G _ _ and that means 1/8*6/7*5/6 = 5/56
Now why do we have to multiply it by 3? Let's say Z & Y were selected out of remaining 6, then Isn't GZY & ZGY & ZYG & GYZ etc mean exactly same? since we are selecting people to form the team and if it's not shouldn't we multiply it by 3! instead of 3.
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From a Group of 8 People, Including George and Nina, 3 people are [#permalink]
Hello everyone,

I understand concept explained in the forum. Nevertheless, while practicing I figured out the following approach:

(1) Probability that George will be selected is 3 out of 8 (as there are 3 places for 8 people).
(2) Afterwards, probability that Nina will not be selected is 5 out of 7 remaining spots (5 because there are 5 people who will not be accepted).

So overall probability is (3*5)/(8*7) = 15 / 56, which is in line with the answer using combinatorics 6C2 / 8C2.

Is the method described above relevant as well?
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Re: From a Group of 8 People, Including George and Nina, 3 people are [#permalink]
a silly way that led me to the answer without actually knowing how to solve it

first, select 3 out of 8 is 3C8=56
then I thought there are 4 scenarios {neither is selected,G is selected,N is selected, both are selected}

I said that each one had the same possibility so 56/4= 14

14/56 is almost 15/56

so I got it right with a wrong procedure
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Re: From a Group of 8 People, Including George and Nina, 3 people are [#permalink]
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