# Probability and statistics

1. Sep 18, 2005

### philipc

I'm not having much luck learning from the examples in my probability book, they only do the most of basic examples, I'm in need of some good examples and solutions, any ideas?

For example here is the problem I'm trying to work on
A car rental agency has 18 compact cars and 12 mid-size cars, if 4 are randomly selected, what is the probability of getting two of each kind?

I'm not sure how to set up the equations
Thanks
Philip

2. Sep 18, 2005

### iNCREDiBLE

What have you tried, my friend? Just use the multiplication rule.

3. Sep 18, 2005

### philipc

4. Sep 18, 2005

### iNCREDiBLE

What Have You Tried?

5. Sep 18, 2005

### iNCREDiBLE

Nvm, here we go. Let us assume that randomly selected means that each of the $\binom{30}{4}$ combinations is equally likely to be selected. Hence the desired proability equals
$$\frac{\binom{18}{2}\binom{12}{2}}{\binom{30}{4}}$$

6. Sep 18, 2005

### philipc

That gives the correct results thank,
my problem I'm still having a hard time following the logic behind picking the numbers to work with.

Here is another example from the book I can't find the logic
If 3 of 20 tires are defective and 4 of them are picked randomly, what is the probability that only one of the defective one will be included.
So I tried this
$$\frac{\binom{17}{3}}{\binom{30}{4}}$$
this would give your total probability of receiving any number of bad tires right?
But the book wants just one bad tire so they use
$$\frac{\binom{3}{1}\binom{17}{3}}{\binom{20}{4}}$$
this is the logic I can't follow, can you explain why they choose these numbers?
Thanks
Philip

7. Sep 19, 2005

### philipc

Got one more lets say P(A|B) = .2 and I know P(B) to be .65, how can I solve for P(A)?

Last edited: Sep 19, 2005
8. Sep 19, 2005

### iNCREDiBLE

There are totally 20 tires (17 proper and 3 defective).

Experiment 1: Pick one defective tire. One (exactly one) defective tire can be picked in $\binom{3}{1}$ different ways.

Experiment 2: Pick three proper tires. This can be done in $\binom{17}{3}$
Using the basic principle of counting we got $\binom{3}{1}\binom{17}{3}$ different ways to pick 1 defective and 3 proper tires among 20.