sm021984 wrote:
A dog breeder currently 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 selected dogs are NOT
littermates ?
A)1/6
B)2/9
c)5/6
D)7/9
E)8/9
There are a total of 9 dogs - 3 have exactly 2 litter mates. 6 have only 1 litter mate.
Say A has two litter mates B and C. Then B also has two litter mates - A and C and C also has two litter mates A and B. There will be only one such group of 3 dogs where each has exactly 2 litter mates.
Of the remaining 6, say D's litter mate is E. Then E's litter mate is D. Neither of them will have any other litter mate.
Similarly, F's litter mate is G and G's litter mate is F.
H's litter mate is I and I's litter mate is H.
A, B, C
D, E
F, G
H, I
In how many ways can we select two dogs such that they ARE litter mates?
You can do it in two ways:
Finding number of combinations:
We can select D, E or F, G or H, I i.e. 3 ways.
Of A, B, C, we can select any two in 3C2 = 3 ways.
Total ways of selecting 2 litter mates is 3+3 = 6
Total ways of selecting 2 dogs out of 9 is 9C2 = 36
Probability of selecting 2 dogs such that they are litter mates = 6/36 = 1/6
Probability of selecting 2 dogs such that they are not litter mates = 1 - 1/6 = 5/6
Directly using probability:
Probability of selecting a dog out of D, E, F, G, H and I = 6/9 = 2/3
Probability of selecting his litter mate = 1/8
Probability of selecting a dog out of A, B and C = 3/9 = 1/3
Probability of selecting his litter mate = 2/8 = 1/4
Probability of selecting 2 litter mates = 2/3 *1/8 + 1/3*1/4 = 1/6
Probability of selecting non litter mates = 1 - 1/6 = 5/6