# Probability question

1. Jan 8, 2009

### Cemre

Let's say we have 10 boxes and I open each of them one by one...

I open the 1st box, there is a toy car in it.
I open the 2nd box, there is also a toy car in it.
I open the 3rd box, there is also a toy car in it.
...
I open the 9th box, there is also a toy car in it. :) wow, I got 9 toy cars in 9 boxes...

what is the probability that 10th box also has a toy car in it?

also generalize 10 to any number...
what is the probability that nth box also has a toy car in it, if all n-1 boxes each have a toy car in them.

2. Jan 8, 2009

### CRGreathouse

There are lots of ways to do this binomial confidence interval problem. One common way is prob = (# successes + 1) / (# trials + 2), which would suggest a 91% chance.

3. Jan 9, 2009

### gmax137

Depends on where you're getting the boxes.

4. Jan 9, 2009

### uart

Yes this is a silly question, unless you give some more information there is no well defined answer.

$$P = \frac{m-9}{n-9}$$