Why Was My First Attempt Wrong in Calculating iPod Probability?

[davedave]

Here is a problem that I found in a library book.

In Canada, 38% of all students have an iPod. What is the probability that 10 out of 20 randomly selected students have AT LEAST ONE iPod?

This is my CORRECT solution.

It is a binomial distribution. So, I do P(having an iPod) = 0.38 which is the success

20 choose 10 times (.38)^10 times (.62)^10 = 0.097

I happened to come up with the correct solution after my first attempt.

In my first attempt, this is what I did.

P(at least one iPod) = 1 - P(no iPod) which gives the wrong answer.


Why is my first attempt wrong? Please explain.

 

[Tedjn]

How did you calculate P(no iPod)?

 

[Klockan3]

First you note down the chance of each configuration to happen, that is just multiply together the probabilities of each event in solitude. Now it doesn't matter which 10 of the 20 students that got ipods, so you got to add together the probabilities of all configurations which have 10 students with ipods. But since all configurations have the same probability you just multiply the probability of one with the number of configurations which is just 20 chose 10.

 

[uart]

In my first attempt, this is what I did.

P(at least one iPod) = 1 - P(no iPod) which gives the wrong answer.

Why is my first attempt wrong? Please explain.

What you most likely did is to misunderstand what is the correct complimentary probability in this case.

Think about it. The complement of "10 out of 20 have and ipod" is definitely not the same thing as "10 our of 20 do not have ipod's"!

So what then is the correct complement? It is that some proportion other than 10 out of 20 (eg 0/20 1/20 2/20 … 9/20 11/20 ... 20/20) have an ipod.

Do you see your mistake now?