A bag has 3 red and 7 blue marbles, two are picked without… Step by Step

Step by step

The question

Question: A bag has 3 red and 7 blue marbles.

Two marbles are picked, one after the other.

The first one is not put back.

What is the probability the first is blue and the second is 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 blue, then red.

That is this path, in red.

3 Multiply along the branches

And means both happen.

We multiply along the path: 7/10, times 3/9.

P(blue then red) = 710 × 39

and: multiply along the branches

Careful!

The first marble is not back.

The second pick is out of 9.

Careful: not 710 × 310 ✗

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

P(blue then red) = 7 × 310 × 9

7 times 3 is 21.

10 times 9 is 90.

So it is 21/90.

P(blue then red) = 2190

4 Simplify

Both 21 and 90 can be divided by 3.

That makes it simpler.

P(blue then red) = 21 ÷ 390 ÷ 3

So it is 7/30.

P(blue then red) = 730

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: 6 plus 21 plus 21 plus 42 is 90.

90/90 is 1.

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

690 + 2190 + 2190 + 4290 = 9090 = 1

Let's look back.

First: independent or not.

Then: draw a tree.

Then: multiply along the branches.

Then: simplify.

Last: check.

Answer