Problem Solving

In a room filled with 7 people, 4 people have exactly 1 sibling in the room and 3 people have exactly 2 siblings in the room. If two individuals are selected from the room at random, what is the probability that those two individuals are NOT siblings?


a 5/21
b 3/7
c 4/7
c 5/7
e 16/21
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Let the people be A B C D E F G

A is a sibling of B
C is a sibling of D
E F G are siblings of each other

So we have these siblings:
A-B
C-D
E-F-G

Number of ways of selecting 2 people from pair 1 (i.e. A-B) = 2C2 = 1

Number of ways of selecting 2 people from pair 2 (i.e. C-D) = 2C2 = 1

Number of ways of selecting 2 people from pair 3 (i.e. E-F-G) = 3C2 = 3

Total number of ways of selecting 2 people out of 7 = 7C2 = 21

Probability of selecting a sibling = 5/21
Thus, probability of not selecting a sibling pair = 1 - 5/21 = 16/21

Answer should be (E)

there are 2 pairs of sibling and 3 sibling

2 people who are sibling can be selected from 2 pairs in 2 ways
2 people who are sibling from 3 sibling are=3c2=3


total possibility of selecting 2 people from 7 people is =7c2=21


probability that two individuals are siblings=(2+3)/21

two individuals are NOT siblings = 1 -5/21=16/21

P = (good outcomes)/(total possible outcomes)

total outcomes = total number of possible pairs that can be made from 7 choices = (7*6)/(1*2) = 21

A good outcome is if we don't choose a sibling pair. Let's count how many siblings pairs we have:

If ABC are all siblings (AB, AC, BC), that gives us 3 siblings pairs that account for the 3 people that have 2 siblings each.
This leaves 2 other sibling pairs, DE and FG, which account for the 4 people that have 1 sibling each.

So we have 3+2=5 sibling pairs.

This means we have 21-5=16 pairs that are NOT made of siblings.

So P(not a sibling pair) = 16/21.

The correct answer is E.