Step by step
The question
Question: A bag has 5 red and 7 blue marbles.
Two marbles are picked, one after the other.
The first one is not put back.
What is the probability both are red?
There are four choices.
First we solve it, the normal way.
Then we pick.
1 Independent or not
The first marble is not put back.
We call them dependent.
So the second pick has one marble fewer: 11.
And the colour picked first has one fewer.
2 Draw a tree
First pick: red is 5 out of 12, blue is 7 out of 12.
After a red: 11 left, and 4 red and 7 blue.
After a blue: 11 left, and 5 red and 6 blue.
We want red, then red.
That is this path, in red.
3 Multiply along the branches
And means both happen.
We get red 5 out of 12 times.
Of those times, 4 out of 11 go on to red.
A part of a part: of means times.
So we multiply along the path: 5/12, times 4/11.
Careful!
The first marble is not back.
The second pick is out of 11.
Before we multiply, we make the numbers smaller.
4 on top and 12 at the bottom both divide by 4.
To multiply fractions, we do top times top, and bottom times bottom.
5 times 1 is 5.
3 times 11 is 33.
So it is 5/33.
4 Check
Let's check the tree.
Each end: multiply along its branches.
We do not cancel, so every bottom is 132.
All the ends together: 20 plus 35 plus 35 plus 42 is 132.
132/132 is 1.
So the tree is right, and so is our path!
5 Pick the choice
Our answer is P(both red) = 5/33.
That is choice A.
6 Why not the others
B is 25/144.
Wrong: 5/12 × 5/12: after one red is taken, 4 red of 11 are left.
That is the Careful mistake from before.
C is 5/12.
Wrong: only the first pick: 5 out of 12.
D is 103/132.
Wrong: 5/12 + 4/11: added instead of multiplying.
Let's look back.
First: independent or not.
Then: draw a tree.
Then: multiply along the branches.
Then: check.
Then: pick the choice.
Last: why not the others.