# Homework Help: Independent events confusion

1. Jan 28, 2012

### synkk

Bag contains 9 blue balls, 3 red balls.
A ball is selected at random, colour recorded and NOT replaced.
A second ball is selected at random, colour recorded.

a) Draw tree diagram (no problem).

Part 2

a) Find probability second ball is red (no problem, 0.25)

b) Find probability both balls selected are red, given that the second ball is red.

I'm having a problem understanding where my logic is wrong.

Using the formula:

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

or

P(both balls are red | second ball is red) = P(both balls are red and second ball is red) / P(second ball is red)

P(second ball is red) = (1/4) or 0.25 (from part a)

P(both balls are red) = (3/12) * (2/11) = (1/22)

Because I need to find:

P(both balls are red and second ball is red)

I do the following to get the intersection

(1/4) * (1/22) = (1/88)

and then divide it by P(second balls is read) to reduce the sample space down to the given event. However this leads to the wrong answer.

Looking at the mark scheme, the just do (3/12) * (2/11) to get the intersection, but following the formula leads me to a different answer.

Looking at the textbook it says that if two events are independent then P(A|B) = P(A), if so then these two events are not independent, and the mark scheme is treating it as if they were independent.

2. Jan 28, 2012

### Ray Vickson

{both are red and second is red} = {both are red} (saying the second is red does not change anything).

RGV

Last edited: Jan 28, 2012
3. Jan 28, 2012

### synkk

So they are independent?

4. Jan 28, 2012

### Ray Vickson

If by "they" you mean the two events {both are red} and {second is red} then NO, they are not independent (one of them is contained in the other). However, that does not matter at all.

RGV

5. Jan 28, 2012

### synkk

By they I mean the events that red or blue is picked. How does it not matter at all? It says on my book here that if two events are independent then P(A|B) = P(A), which is essentially what is going on here, but they are dependent, which is whats confusing me.

6. Jan 28, 2012

### Ray Vickson

All the events in your first posting involved only red balls, not blue ones. Of course, either a red one or a blue one is picked: nothing else can possibly happen! However, that is not the issue. You originally said you wanted P{both are red and second is red}, and you claimed to be confused about having to compute P{A & B} for A and B possibly not independent; in the context of your post you were taking A = {both are red} and B = {second is red}. Again, I repeat: they are dependent, but it does not matter. Maybe you are just trying to apply some formulas without thinking; I suggest you look more carefully at the situation.

RGV

7. Jan 29, 2012

thank you