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?
5/21
3/7
4/7
5/7
16/21

OA E

IMO C
reqd probability is nothing but to select one out of those 4 who have 1 siblings and to select one out of those 3 who have 2 siblings
it can be done in 4C1 and 3C1 ways
total two can be selected from 7 in 7C2 ways
p=4C1*3C1/7C2=4/7
Pls confirm the ans

Answer.: E

Let the 7 people be denoted by A1, A2, B1, B2, C1, C2 and C3 where "same letter" means "brotherhood", for instance: A1 has exactly one sister/brother (A2) and C3 has exactly two (C1 and C2).

Without taking into account the order of choice of each person chosen to be present in a pair of people (from now on), there is a total of C(7,2) = 21 available pairs of people and, from them, only 5 choices (pairs) do involve "brotherhood":

A1 with A2, B1 with B2, C1 with C2, C1 with C3 and C2 with C3.

That means that 21-5 = 16 choices are pairs of people that are NOT siblings (to each other in the same pair) and we get the answer: 16/21.

Let ABCDEFG be the people in room

4 has excatly 1 sibling
AB and BC
now other have 2 siblings
DE EF and EG

Total 7C2----7*6/ 2 =21

Chances that they are siblings
5 cases

so probablity 5/21

Probablity of Not beingsibling=(1-probablity that they are sibling)
1-5/21=16/21
so
E matches

Total ways of pick 2 from 7: 7P2 = 7*6 = 42

To pick group that those two individuals are NOT siblings

For the 4 that have exactly one sibling:

Pick 1 from that 4, say Jack = 4P1
then pick some one else that is not Jack's sibling = 5P1 ( Why 5P1, after picking Jack, you have 6 left. Among those 6, 1 needs to be eliminated because it is Jack's sibling)

No. of ways = 4*5 = 20

For the 3 that have exactly two sibling:

Pick 1 from that 3, say Kelvin = 3P1
then pick some one else that is not Kelvin's sibling = 4P1 ( Why 4P1, after picking Jack, you have 6 left. Among those 6, 2 need to be eliminated because they are Jack's siblings)

No. of ways = 3*4 = 12

Probability = (20+12)/42 = 16/21

Then Pick E

shovan85 wrote: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?
5/21
3/7
4/7
5/7
16/21

Let's say that the 7 people are ABCDEFG.

4 people have exactly 1 sibling:
Let's say that A and B are siblings and that C and D are siblings.
This means:
A has 1 sibling (B).
B has 1 sibling (A).
C has 1 sibling (D).
D has 1 sibling (C).

3 people have exactly 2 siblings:
Let's say that E, F and G are all siblings of each other.
This means:
E has 2 siblings (F and G).
F has 2 siblings (E and G).
G has 2 siblings (E and F).

Total number of sibling pairs = 5: AB, CD, EF, EG, FG.
Total number of pairs that can be formed from 7 people: 7C2 = 21.
P(sibling pair) = 5/21
P(not sibling pair) = 1 - 5/21 = 16/21.

The correct answer is E.