There are 9 people in the room. There are two pairs of

another ways:

1. \(p=\frac{C^9_2-C^2_1}{C^9_2}=1-\frac{2}{36}=\frac{17}{18}\)

or

2. \(p=\frac{P^9_2-P^2_1*P^2_2}{P^9_2}=1-\frac{4}{72}=\frac{17}{18}\)

or

3. \(p=\frac49*\frac78+\frac59*\frac88=\frac{68}{72}=\frac{17}{18}\)

probability questions always have a few ways

bmwhype2, thanks for your question
+1

= 1- (2C2*2/9C2) = 1-1/18 = 17/18

Method 1 : total number of ways of choosing 2 people out of 9 = 9C2 = 36

probability of choosing two people so that they are siblings = 2C1 = 2 { since there are two pairs of siblings }

required probability = (36 -2) /36 = 17/18.

Method 2 : total number of ways = 36

let say the nine people that we have are A, B, C, D, E, S11, S12, S21, S22 where (S11, S12) and (S21, S22) are sibling pairs.

number of ways one can choose two people is :

choose 2 from A, B, C, D, E . Number of ways = 5C2 = 10.

choose 1 from A, B, C, D, E AND 1 from (S11, S12, S21, S22) in 5*4 = 20 ways

choose 1 from (S11, S12) and 1 from (S21, S22) in 4 ways .

total number of ways = 34

probability = 34/36 = 17/18

There are 9 people in the room. There are two pairs of siblings within that group. If we choose two people, what is the probability that they will not be siblings?

Soln: probability that they will not be siblings
= 1 - Probability that they will be siblings
= 1 - (4/9 * 1/8)
= 1 - (1/18)
= 17/18

Hi Walker,

With reference to your solution for the married couples problem where in we have to select 4 people out of 6 couples such that none is married to each other,can this problem be tackled in the same way?


How will we arrive at the solution?

Thanks!

regards,

You can use the same approach but it's important to note that not all people have siblings.

\(C^5_2\) - the number of options to choose 2 people out of 5 people without siblings
\(C^5_1*C^2_1*C^2_1\) - the number of options to choose 1 person out of 5 without siblings, choose 1 pair of siblings out of 2, choose 1 person out of the pair.
\(C^2_1*C^2_1\) - the number of options to choose 1 person out of 2 for each pair of siblings.

\(p = \frac{C^5_2+C^5_1*C^2_1*C^2_1+C^2_1*C^2_1}{C^9_2} = \frac{10+20+4}{36} = \frac{34}{36} = \frac{17}{18}\)

My approach : 1- p(1 minus the probability that the people we pick will be siblings)
We have 9 people:

A1 - A2 - B1 - B2 - C - D - E - F - G

Probability of picking siblings:
If we first pick A1, Probability of picking A1 \(p=\frac{1}{9}\frac\) (we pick 1 between 9)

Then, probability of picking the sibling of A1, A2, after having picked A1: \(p=\frac{1}{8}\frac\) (we pick 1 between the remainder 8 people) * \(\frac{1}{9}\frac\) (conditional probability of picking A2 after having picked A1) = \(\frac{1}{72}\frac\)

But, in addition, we could have picked A2 and then A1. Then: \(p=\frac{1}{72}\frac\) *2 = \(\frac{2}{72}\frac\) = \(\frac{1}{36}\frac\)

The same if we picked first B1 and secondly B2 or ("or" means "+") if we first picked B2 and secondly B1. Then, the total probability of picking siblings is: \(p=\frac{1}{36}\frac\) * 2 = \(\frac{2}{36}\frac\) = \(\frac{1}{18}\frac\)

Then, 1 - p = \(\frac{18}{18}\frac\) - \(\frac{1}{18}\frac\) = \(\frac{17}{18}\frac\)

Is right this logic? I am understanding the problem correctly

Absolutely, though not the most elegant way of solving.

There are 9 people in the room. There are two pairs of siblings within that group. If we choose two people, what is the probability that they will not be siblings?
The reversal probability approach is the most efficient way to solve. But I explore the other methods just for practice. Tell me your opinions, I have a little doubt on the reversal combinatorial approach.

Probability approach:

XXXXXAABB

\(P= \frac{6}{9} * \frac{1}{2}*2 +\frac{5}{9}*\frac{4}{8} = \frac{12}{18} + \frac{5}{18} = \frac{17}{18}\)

\(\frac{6}{9}*\frac{1}{2}*2\): we choose 6 people out of 9 (XXX XXA, 6/9) and then 1 people out of the other 2 (BB 1/2). Then multiply by 2 because we choose again 6 people out of 9 (XXXXXB) and then 1 people out of the other 2 (AA).

\(\frac{5}{9}*\frac{4}{8}\): then we focus only on the other 5 that are not siblings (XXXXX). We choose this 5 people (XXXXX) out of all 9 (then 5/9) and we combine this one with the remaining 4 of the group (XXXX) of the remaining 8.

Reversal Probability approach:
P = 1-q (q=probability of choosing one sibling)

\(q= \frac{2}{9} * \frac{1}{8} + \frac{2}{9} * \frac{1}{8} = \frac{4}{72} = \frac{1}{18}\)
\(P = 1-q = \frac{18}{18} - \frac{1}{18} = \frac{17}{18}\)

\(\frac{2}{9}\): probability of picking the person A from AABBXXXXX
\(\frac{1}{8}\): probability of picking the other A from the remaining ABBXXXXX

Then we add the probability of picking B from AABBXXXXX and probability of picking again B from AABXXXXX

Combinatorial approach:

\(P= \frac{{C^6_1 * C^2_1 + C^6_1 * C^2_1 + C^5_2}}{C^9_2}= \frac{{6*2 + 6*2 + 10}}{36} = \frac{34}{36} = \frac{17}{18}\)

\(C^6_1 * C^2_1\): we choose one people from AXXXXX and combine with one people of BB
\(C^6_1 * C^2_1\): we pick one people of BXXXXX and combine it with one people of AA
\(C^5_2\): we choose 2 people from XXXXX
\(C^9_2\): total combinations of 2 from 9.

Reversal Combinatorial approach:

\(q= \frac{{C^2_1 + C^2_1}}{C^9_2}= \frac{{2 + 2}}{72} =\frac{4}{72} = \frac{1}{18}\)
\(P = 1-q = \frac{18}{18} - \frac{1}{18} = \frac{17}{18}\)

\(C^2_1\): we choose 2 persons from the sibling (AA)
\(C^2_1\): we choose 2 persons from the sibling (BB)

Can someone please explain where i'm going wrong?

I'm using the approach: No Sibling Pair = 1 - Probability of sibling pair.

P = \(\frac{(9C4)(1C1)}{(9C2)}\) -- why is this wrong? Is this equation not implying that at first, I have the probability of choose 4 people out of 9, then I can only pick 1 so my top is the favorable solution. My denominator is the total number of outcomes.

The numerator is wrong. We are not choosing 4 people, we are choosing 2.

How many ways to choose a sibling pair from 9 people where there are two pairs of siblings (X, X, Y, Y, A, B, C, D, E)? Only two ways XX and YY.

Hence, P = 1 - 2/(9C2) = 1 - 1/18 = 17/18.

There are 9 people in the room. There are two pairs of siblings within that group. If we choose two people, what is the probability that they will not be siblings?

I need to choose two people from a group of 9 . I will choose 1 the probablitiy is 100% (9/9), then I have to choose the second person in such a manner that he is not the sibling of the 1 I chose, therefor amongst the 8 I am still left with, the probability becomes 7/8 (excluding the one sibling of the first). Please correct my approach and preferably donot provide me with an alternative solution. It would be of great help.

The point is that if you choose a person which has no sibling then the probability that the second person won't be his/her sibling would be 1. That's why this approach does not work.

Correct approach is here: there-are-9-people-in-the-room-there-are-two-pairs-of-58609.html#p1357845

Hope it helps.

Hi Bunuel,
I was solving this as follows (similar to walker's solution #3):

P(siblings are not chosen) = P(both are picked from 5-singles) + P(one is picked from 5-singles and one from 4-siblings) + P(one is picked from 4-siblings and other is picked from remaining sibling)
= 5/9*4/8 + 5/9*4/8 + 4/9*2/8
=12/18 = 2/3

How is this logic flawed?

You could also think of it like this:

Sibling Pair 1: A1, A1
Sibling Pair 2: B1, B2

2 People Selected but both not siblings would mean, the sum of these cases
Case 1: A1 selected but not A2 = 1/9 * 7/8
Case 2: A2 selected but not A1 = 1/9 * 7/8
Case 3: B1 selected but not B2 = 1/9 * 7/8
Case 4: B2 selected but not B1 = 1/9 * 7/8
Case 5: Any one other than the 4 siblings selected on the first go followed by anyone from the remaining 8 selected on the second go = 5/9 * 1

FInal = 4 * (1/9 * 7/8) + 5/9 = 17/18