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Probability

by abhirup1711 » Wed Jun 12, 2013 6:26 am
A dog breeder has 9 breeding dogs. 6 of the dogs have exactly 1 littermate, and 3 of the dogs have exactly 2 littermates. If 2 dogs are selected at random, what is the probability that both the selected dogs are NOT littermates?
1/6
2/9
5/6
7/9
8/9
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by Brent@GMATPrepNow » Wed Jun 12, 2013 7:06 am
abhirup1711 wrote:A dog breeder has 9 breeding dogs. 6 of the dogs have exactly 1 littermate, and 3 of the dogs have exactly 2 littermates. If 2 dogs are selected at random, what is the probability that both the selected dogs are NOT littermates?
1/6
2/9
5/6
7/9
8/9
Littermates are dogs born in the same batch (litter).

Here's the counting approach.

Let the dogs be represented by the letters A to I.
The following meets the given conditions.
- A and B are littermates
- C and D are littermates
- E and F are littermates
- G, H and I are littermates

We want to find P(selected dogs are not littermates)
Let's use the complement.
So, P(selected dogs are not littermates) = 1 - P(selected dogs are littermates)

P(selected dogs are littermates)
How many possible outcomes satisfy the condition that the two dog ARE littermates?
We have:
- A and B
- C and D
- E and F
- G and H
- G and I
- H and I
There are 6 possible outcomes.

In how many ways can we select 2 dogs from 9 dogs?
Since order doesn't matter, we can select the dogs in 9C2 ways (36 ways).

So, P(selected dogs are littermates) = 6/36 = 1/6


P(selected dogs are not littermates) = 1 - P(selected dogs are littermates)
= 1 - 1/6
= [spoiler]5/6 [/spoiler]
= C

Cheers,
Brent

Aside: If anyone is interested, we have a free video on calculating combinations (like 9C2) in your head: https://www.gmatprepnow.com/module/gmat-counting?id=789
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by GMATGuruNY » Wed Jun 12, 2013 7:16 am
I posted a solution to this problem here:
https://www.beatthegmat.com/dog-breeder-t134956.html

Check here for a similar problem:
https://www.beatthegmat.com/siblings-t66114.html
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by Brent@GMATPrepNow » Wed Jun 12, 2013 7:30 am
abhirup1711 wrote:A dog breeder has 9 breeding dogs. 6 of the dogs have exactly 1 littermate, and 3 of the dogs have exactly 2 littermates. If 2 dogs are selected at random, what is the probability that both the selected dogs are NOT littermates?
1/6
2/9
5/6
7/9
8/9
Here's the probability approach.

Let the dogs be represented by the letters A to I.
The following meets the given conditions.
- A and B are littermates
- C and D are littermates
- E and F are littermates
- G, H and I are littermates

We want to find P(selected dogs are not littermates)

For the probability approach, it helps to say that one dog is selected first and the other dog is selected second.
Notice that there are two different ways in which the two dogs are NOT littermates:
#1) 1st dog is from one of the 2-dog pairings (AB, CD, or EF) and 2nd dog is not a littermate
#2) 1st dog is from the 3-dog group (GHI) and 2nd dog is not a littermate

So, . . .
P(selected dogs are not littermates) = P(1st is from a 2-dog pairing and 2nd is not a littermate OR 1st is from the 3-dog group and 2nd is not a littermate
= P(1st is from a 2-dog pairing and 2nd is not a littermate) + P(1st is from the 3-dog group and 2nd is not a littermate)
= (6/9)(7/8) + (3/9)(6/8)
= 42/72 + 18/72
= 60/72
=[spoiler] 5/6[/spoiler]
= C

Cheers,
Brent
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