Solving Probability Questions: 2 Homes, Cats, Dogs

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In summary, the question is about the probability of two homes with a specific combination of pets (2 cats and 1 dog) and another two homes with a different combination (3 dogs). The total number of possibilities is 12C4, and the probability of picking a cat is 1/3 and a dog is 2/3, with no replacement after picking an animal. The order does not matter and there is no other known information. The question could be solved by dividing the subset of the population into specific groups. The probability would not change based on the number of cats and dogs in the population. The probability of one home picking cat-cat-dog is unknown and more information is needed to solve the question.
  • #1
miscellaneous_
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Hey, I think this is a combinatorial question and I'm too far removed from stats to remember. I thought it was originally a binomial distribution but then that didn't make sense after I calculated it.

What is the probabilty of 2 homes with 2 cats and 1 dog and 2 homes with 3 dogs?

How would I solve this type of question in general. Thanks.
 
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  • #2
Welcome to PF;
The question cannot be answered without more context.
What is the known information?
 
  • #3
Thanks Simon,
There are 4 cats and 8 dogs in total.
The probability of picking a cat is simply 1/3 and the probability of picking a dog is 2/3 initially (i.e. one animal is not inherently more probable to be chosen).
There is no replacement after picking an animal.
Order does not matter.
There is nothing else that I can think of. The question could be anything, 2 teams with 2 boys and 1 girl and 2 teams with 3 girls. Basically dividing the subset of the population into specific groups.
Is there anything specific I'm not thinking about because I feel like a moron.
I hope there is a really complex answer ... it's probably not.

The total number of possibilities is 12C4 right?
 
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  • #4
OK - so houses are twice as likely to want a dog as a cat.

Guessing:
There are two homes - they take turns picking until each has three pets and each pick is independent?

If the guess is true, then:
You cannot run out of one animal for the selections of interest, so it does not matter how many dogs and cats are in the population: the probabilities won't change.

Each home with combination cat-cat-dog (any order)
What is the probability of one home picking cat-cat-dog?
 
  • #5


I would approach this question by first clarifying the assumptions and parameters involved. For example, are we assuming that the homes are randomly selected or are they specific homes with known pet ownership? Are we considering all possible combinations of pets or only those that meet the given criteria?

Once these details are established, we can use principles of combinatorics and probability to solve the question. In this case, we can use the formula for calculating the probability of a specific combination of events (in this case, 2 homes with 2 cats and 1 dog and 2 homes with 3 dogs) out of all possible combinations.

In general, solving probability questions involves identifying the relevant variables, determining the probability distribution, and then using mathematical tools such as combinatorics, binomial distribution, or other statistical methods to calculate the desired probability. It is also important to keep in mind any assumptions or limitations in the problem and to carefully interpret the results in the context of the problem.
 

FAQ: Solving Probability Questions: 2 Homes, Cats, Dogs

1. What is the probability of choosing a home with at least one cat or dog?

The probability of choosing a home with at least one cat or dog depends on the total number of homes and the number of homes that have either a cat or a dog. To calculate this probability, divide the number of homes with at least one cat or dog by the total number of homes.

2. What is the probability of choosing a home with both a cat and a dog?

The probability of choosing a home with both a cat and a dog depends on the total number of homes and the number of homes that have both a cat and a dog. To calculate this probability, divide the number of homes with both a cat and a dog by the total number of homes.

3. If a home has a cat, what is the probability that it also has a dog?

This question can be answered by calculating the conditional probability of a home having a dog given that it has a cat. To calculate this, divide the number of homes with both a cat and a dog by the number of homes with a cat.

4. What is the probability of choosing a home with neither a cat nor a dog?

The probability of choosing a home with neither a cat nor a dog depends on the total number of homes and the number of homes that have neither a cat nor a dog. To calculate this probability, divide the number of homes with neither a cat nor a dog by the total number of homes.

5. How does the number of homes with cats and dogs affect the overall probability of choosing a home with at least one pet?

The more homes with cats and dogs there are, the higher the probability of choosing a home with at least one pet. This is because there are more options to choose from, increasing the chances of selecting a home with a cat or a dog. However, this probability also depends on the total number of homes, as well as the distribution of cats and dogs among the homes.

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