Question 1083529: Hello, I'm having problems with these and I need sort of explaination on how to solve them. Any help or a point in the right direction would be much appreciated. Thank you in advance for your time and help
Michael and Angela run an animal shelter that houses only cats and dogs. The ration of cats to dogs is 5:4. Of the cats, 40% are kittens, while 3/7 of the dogs are puppies. The shelter houses a total of 63 animals.
1.) A group of 6 animals is chosen at random from the shelter. What is the probability that the group contains at least one kitten? Give your answer to 4 decimal places.
2.) A group of 8 adult animals is chosen at random from the shelter. What is the most likely number of cats and dogs in this group and what is the probability of that arrangement happening? Give your probability to 4 decimal places.
___ cats and ___ dogs with a probability of _____
3. A group of 10 baby animals is chosen from the shelter. Is the probability distribution for the makeup of this group symmetric or skewed? Justify your answer.
Answer by mathmate(429)   (Show Source):
You can put this solution on YOUR website! You are probably working on hypergeometric distribution.
We will start with a contingency diagram, which breaks down the cats and dogs (C and D), and adults and babies (A, B)
--- C D Tot
A 21 16 37
B 14 12 26
Tot 35 28 63
1. 6 animals chosen from 63.
P(#kittens>=1)
=1-P(#kittens=0)
=
=0.7942
But that's not the only way. Using hypergeometric distribution, the probability
P(#kittens=0)=C(14,0)*C(49,6)/(63,6)
where C(14,0) is the number of combinations of 0 objects taken from 14.
The formula is basically
C(S,s)*C(F,f)/C(S+F,s+f)
where
S=number of successes in pool
s=number of successes in sample
F=number of failures in pool
f=number of failures in sample.
Here there are 14 kittens in pool (S=14) and 49 others (F=49)
we are calculating no kittens, so s=0, and f=6.
This evaluates also to 0.7942.
We will be using this formula to solve the remaining two problems.
2. 8 animals (s+f=8) from 37 adults (S+F=37).
The distribution of cats and dogs is 21:16, so the most likely arrangements could be
6:2, 5:3, 4:4
We'll see.
P(6:2)=C(21,6)*C(16:2)/(37,8)=0.1687
P(5:3)=C(21,5)*C(16:3)/(37,8)=0.2952
P(6:2)=C(21,4)*C(16:4)/(37,8)=0.2822
Which shows that the most probable arrangement is 5 cats and 3 dogs.
3. A group of 10 baby animals (out of 14 kittens and 12 puppies) is chosen. The probability distribution is skewed. You can show this by showing that P(4:6) does not equal P(6:4), or other similar ratios.
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