rickyishere 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?

a) 5/21
b) 3/7
c) 4/7
d) 5/7
e) 16/21

I was looking at the solution given for the CAT question and still dont understand how siblings were counted .Can some one please help?

Thanks

this one has been discuss for a number of times
here is the net output

As there are 4 people with exactly 1 sibling each: we have two pairs of siblings (1-2; 3-4).
As there are 3 people with exactly 2 siblings each: we have one triple of siblings (4-5-6).

Solution #1
# of selections of 2 out of 7 7C2=21
# of selection 2 people which are not siblings=2C1*2C1(one from first pair of siblings*one from second pair of siblings)+2C1*3C1(one from first pair of siblings*one from triple)+2C1*3C1(one from second pair of siblings*one from triple)=4+6+6=16
P=16/21

Solution #2
# of selections of 2 out of 7 7C2=21
# of selections of 2 siblings=3C2+2C2+2C2=3+1+1=5
P=1-5/21=16/21

Solution #3
2*3/7*4/6+2*2/7*2/6=4/7+4/21=16/21.

rickyishere 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?

a) 5/21
b) 3/7
c) 4/7
d) 5/7
e) 16/21

I was looking at the solution given for the CAT question and still dont understand how siblings were counted .Can some one please help?

Thanks

No of ways in which 2 people can be selected from a group of 7 = 7C2 = 21
No of ways in which the two people selected are siblings = 2*2C2 + 3C2 = 5

Required probability = (21-5)/21 = 16/21

I just cant get the logic. Could someone please explain again in detail.
What i did do was

(4C1*1C1 +3C1*2C1)/7C2

If you pick one fellow with 1 sibling so he has just one choice left so 4C1*1C1
If you pick a person with 2 siblings so he has two choices left so 3C1*2C1

Some do clarify

Thanks in advance

E

eaakbari wrote:I just cant get the logic. Could someone please explain again in detail.
What i did do was

(4C1*1C1 +3C1*2C1)/7C2

If you pick one fellow with 1 sibling so he has just one choice left so 4C1*1C1
If you pick a person with 2 siblings so he has two choices left so 3C1*2C1

Some do clarify

Thanks in advance

E

If you pick one fellow with 1 sibling (It can be done in 2 ways) You can pick the second fellow in 5 ways
If you pick a person with 2 siblings (It can be done in 3 ways) you can pick the second fellow in 4 ways

Let X1 X2 Y1 Y2 Z1 Z2 Z3 be the siblings

X1, Y2
X1, Y1
X1, Z1
X1, Z2,
X1, Z3
X2, Y2
X2, Y1
X2, Z1
X2, Z2,
X2, Z3
Are the choices for the first condition

Z1, X1
Z1, X2
Z1, Y1
Z1, Y2
Z2, X1
Z2, X2
Z2, Y1
Z2, Y2
Z3, X1
Z3, X2
Z3, Y1
Z3, Y2

Are the choices for second condition
X1, Z1
X1, Z2,
X1, Z3
X2, Z1
X2, Z2,
X2, Z3
Are common in both, so the required number = 10+12-6 = 16