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

Similar question to practice: from-a-group-of-8-volunteers-including-andrew-and-karen-160883.html

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 \(\frac{3*6*5}{8*7*6}\) = \(\frac{3*5}{8*7}= \frac{15}{56}\) .

Answer:C

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 = \(\frac{Number of outcomes with A and not B }{Total number of outcomes}\)

= \(\frac{15 }{56}\)

Bunuel, Karishima, someone please!

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 ?

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.

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..

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).

Thank you for your help!

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 !!

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

Answer: C

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

Can we use fundamental counting principle to achieve the no of ways for selecting 3 persons from 8 with George included & Nina excluded. Bunuel KarishmaB

I used same & got 6x5 = 30 ways, please help me figure out the gap/flaw in my approach.

Thanks in advance
Deepak

6 * 5 counts both X-Y and Y-X pairs as distinct, even though the order doesn’t matter here. You should divide that number by 2 to avoid double-counting.

FCP arranges the outcomes for you which means that if you want to just "select" you must un-arrange. As Bunuel suggested, you need to divide by 2 to un-arrange (just like to arrange, you would multiply by 2)

Check this video for discussion on FCP: https://youtu.be/LFnLKx06EMU