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
d. 5/7
e. 16/21

I don't fully understand the explanation MGMAT gave. Any help would be greatly appreciated.

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?



seven people in hte room are A,B,C,D,E,F,G

4 people have exactly 1 sibling in the room => A,B,C,D ( A and B are siblings , C and D are siblings)

3 people have exactly 2 siblings in the room => E,F,G (E, F and G are siblings).

what is the probability that those two individuals are NOT siblings?

pick 1 from A,B and pick 1 from C, D (or) pick 1 from C, D and pick 1 from E,F,G (or) pick 1 from E,F,G or pick 1 from A,B

1/2*1/2 + 1/2 * 1/3+ 1/3 * 1/2 = 1/4 + 1/6 + 1/6 = 7/12.


PS: Its been more than 10 I had any touch with probability. sorry if i have miss lead you...i tried my best to answer this.

jscpba 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 don't fully understand the explanation MGMAT gave. Any help would be greatly appreciated.

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.

What is the quickest way to come to total number of pairs in a group of x people. In this instance it was 21, and I can obviously come up with that, but what is the shortcut or formula?

The MGMAT explanation says use:

(7*6)/2 to come to 21 combinations

Where do the 7 and 6 come from?

jscpba wrote:The MGMAT explanation says use:

(7*6)/2 to come to 21 combinations

Where do the 7 and 6 come from?

nCr = n!/(r!)(n-r)!

So, 7C2 = 7!/(2!)(7-2)! = 7!/(5!)*(2!) = (5!)*6*7/(5!)*2! (Since, 7! = 5!*6*7)
= 6*7/2 = 21.

I see two different answers.. Please tell the correct answer

nice

so there is a set of 3 siblings (those who have exactly 2)
+ 2 other sets of siblings (exactly 1)

3 + 2 + 2 ---> total 7

2 of the set of 3 can be selected in 3C2 = 3
2 of 2 from second set in 1 way
2 of 2 from third set in 1 way

Total ways to select is 7C2 = 21

prob of selecting 2 siblings = 5/21

so prob of not selecting is 1 - prob of selecting 2 = 1 - 5/21 = 16/21

man i'm having the worst time wrapping my head around probability and combinatorics! anyone have a good resource to help with this stuff? i've gone through the mgmat literature on the topics twice and still not fully getting it!

aleph777 wrote:man i'm having the worst time wrapping my head around probability and combinatorics! anyone have a good resource to help with this stuff? i've gone through the mgmat literature on the topics twice and still not fully getting it!

Go through NOVA book. It is good.

aleph777 wrote:man i'm having the worst time wrapping my head around probability and combinatorics! anyone have a good resource to help with this stuff? i've gone through the mgmat literature on the topics twice and still not fully getting it!

This seems to be oft recommended:

https://www.amazon.com/gp/product/193624 ... PDKIKX0DER

Good luck!