Conditional Probability - Teacher says I'm wrong

[cdm1a23]

Hi Everyone,

Let's see if someone here can do a better job than my teacher! I have one of the least helpful stat teachers ever. She told me that I was wrong about the following problem. I am not saying that she is wrong or right, but when I asked her to explain why I was wrong, she told me that I should just use the formula, and she couldn't explain it until I had taken higher level classes. Here is the problem:

Each person owns one and only one of the following pets:

Dog = 1699 people
Cat = 1418 people
Bird = 826 people
Fish = 504 people
Lizard = 233 people
Other = 177 people

Total People = 4857
Total Pets = 4857

A = Event that a random person owns a Bird or a Fish
B = Event that a random person owns a Dog, a Cat or a Bird

Given that a randomly selected person owns a Bird or a Fish, what is the probability a randomly selected person owns a dog, a cat, or a bird?

I said that the answer would be Bird / (Bird + Fish) = 826/1330 = .621, because we know that the Randomly selected person owns a bird or a fish, and bird is the only shared result of A and B. My teacher said this was wrong, and a formula had to be used to get the right answer, but she refused to go into more depth??? Can someone here please explain what my mistake was?

Thanks very much for your help.

[whatta]

I said that the answer would be Birds / (Birds + Fish) = 826/1330 = .621, because we know that the Randomly selected person owns birds or fish...i dont think it means "randomly selected person will always own either bird or fish". you have probably missed something in the problem definition.

[Moo Of Doom]

You seem to think that pets are exclusive, but surely someone could own fish AND a dog.

EDIT: Hmm... although without more information, I don't think it's solvable if not interpreted your way. What was the answer?

[whatta]

I think this (http://en.wikipedia.org/wiki/Bayes_formula#Example_.231:__Conditional_probabili ties) is the formula you teacher envisions.

[cdm1a23]

You seem to think that pets are exclusive, but surely someone could own fish AND a dog.

EDIT: Hmm... although without more information, I don't think it's solvable if not interpreted your way. What was the answer?

Hi,

Just got back from class. The given answer was .63... Did anyone get their own solution for this?

Thanks Again.

[verty]

I don't see how this can work because if a random person necessarily owns either a bird or a fish or both, that limits the number of people to about 1300, not enough to account for the 1700 who own dogs.

[cdm1a23]

hi everyone... assume one pet type per owner... i'll change the original question to make it better

[jing]

cdm1a23 I see nothing wrong with your statement of the problem. You made it clear that no person owns more than one pet.

You implied that the teacher had given you a formula but you never stated what it was. Was it the conditional probability formula?

using B|A to mean the event B given that the event A has occured

(A and B) to mean both the event A and the event B has occured

P(A) to mean the probability of A occuring

then the formula for conditional probability is

P(B|A) = P(A and B)/P(A)

P(A) = (826+504)/4857 = 1330/4857

(A and B) in your case is the event the person chosen at random owns a bird

The sample space for A is the set {bird owners, fish owners}
The sample space for B is the set {dog owners, cat owners, bird owners}
The sample space for (A and B) is the set {bird owners}

So P(A and B) = 826/4857

So P(B|A) = P(A and B)/P(A) = 826/4857 div 1330/4857

= 826/4857 X 4857/1330 = 826/1330 = .621 to 3 decimal place

So what formula did your teacher use to get .63?
How did she arrive at this answer?

[ssd]

I said that the answer would be Bird / (Bird + Fish) = 826/1330 = .621, because we know that the Randomly selected person owns a bird or a fish, and bird is the only shared result of A and B. My teacher said this was wrong, and a formula had to be used to get the right answer, but she refused to go into more depth??? Can someone here please explain what my mistake was?



What you have done is right.

[whatta]

What you have done is right.something relevant (http://content.ytmnd.com/content/a/2/2/a224551a525ebf5e30cedf1f4ae16b99.gif).

[cdm1a23]

cdm1a23 I see nothing wrong with your statement of the problem. You made it clear that no person owns more than one pet.

You implied that the teacher had given you a formula but you never stated what it was. Was it the conditional probability formula?

using B|A to mean the event B given that the event A has occured

(A and B) to mean both the event A and the event B has occured

P(A) to mean the probability of A occuring

then the formula for conditional probability is

P(B|A) = P(A and B)/P(A)

P(A) = (826+504)/4857 = 1330/4857

(A and B) in your case is the event the person chosen at random owns a bird

The sample space for A is the set {bird owners, fish owners}
The sample space for B is the set {dog owners, cat owners, bird owners}
The sample space for (A and B) is the set {bird owners}

So P(A and B) = 826/4857

So P(B|A) = P(A and B)/P(A) = 826/4857 div 1330/4857

= 826/4857 X 4857/1330 = 826/1330 = .621 to 3 decimal place

So what formula did your teacher use to get .63?
How did she arrive at this answer?


The only formula we have used in class is the one you used. I now believe the teacher just made a mistake, and then blew me off when I tried to explain my method and why I didn't choose .63 on the exam. Thanks for writing it out, and showing me that someone else was getting .621 also.

[cdm1a23]

Thanks for the additional inputs everyone. Whatta... LOL at the letter. I'm not sure what my reaction would be to that if my kid had that sent home with them. Thanks SSD for the confirmation.

[colin.mcenroe]

It's just another brick in the wall.