A bag has 4 red and 6 blue marbles, two are picked without… Step by Step

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The question

Question: A bag has 4 red and 6 blue marbles.

Two marbles are picked, one after the other.

The first one is not put back.

What is the probability both are red?

1 Independent or not

The first marble is not put back.

1st → changes the 2nd

We call them dependent.

dependent:

the 1st changes the 2nd

So the second pick has one marble fewer: 9.

And the colour picked first has one fewer.

2nd pick = 10 − 1 = 9 marbles left

2 Draw a tree

We want red, then red.

That is this path, in red.

3 Multiply along the branches

And means both happen.

We multiply along the path: 4/10, times 3/9.

P(both red) = 410 × 39

and: multiply along the branches

Careful!

The first marble is not back.

The second pick is out of 9.

Careful: not 410 × 410 ✗

To multiply fractions, we do top times top, and bottom times bottom.

P(both red) = 4 × 310 × 9

4 times 3 is 12.

10 times 9 is 90.

So it is 12/90.

P(both red) = 1290

4 Simplify

Both 12 and 90 can be divided by 6.

That makes it simpler.

P(both red) = 12 ÷ 690 ÷ 6

So it is 2/15.

P(both red) = 215

5 Check

Let's check the tree.

Each end: multiply along its branches.

We do not cancel, so every bottom is 90.

All the ends together: 12 plus 24 plus 24 plus 30 is 90.

90/90 is 1.

So the tree is right, and so is our path!

1290 + 2490 + 2490 + 3090 = 9090 = 1

Let's look back.

First: independent or not.

Then: draw a tree.

Then: multiply along the branches.

Then: simplify.

Last: check.

Answer