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A bag has 6 green and 4 yellow marbles, two are picked without…

Step by step, with the reason for every step · Conditional probability · Probability

Step by step

The question

Question: A bag has 6 green and 4 yellow marbles.

Two marbles are picked, one after the other.

The first one is not put back.

What is the probability the second is green, given the first is yellow?

The line means given.

We already know the first is yellow.

P(2nd green | 1st yellow) = ?

P stands for probability.

The line stands for given.

P = probability, | = given

1 Draw a tree

2 Go along the given branch

Given the first is yellow: we go along the yellow branch only.

3 Read the next branch

From there, green is the branch with 6/9.

So the probability is 6/9.

One yellow is already out of the bag.

P(2nd green | 1st yellow) = 69

Careful!

That is the probability of both.

Here the first is already yellow: we only need the second branch.

Careful: not 410 × 69 ✗

4 Simplify

Both 6 and 9 can be divided by 3.

That makes it simpler.

P(2nd green | 1st yellow) = 6 ÷ 39 ÷ 3

So it is 2/3.

P(2nd green | 1st yellow) = 23

5 Check

Let's check with the rule: the probability of both, divided by the probability of the first being yellow.

P(both) ÷ P(1st yellow) = (410 × 69) ÷ 410

Times 4/10, then divided by 4/10: we are back to 6/9.

The same.

So the answer is right!

= 69

Let's look back.

First: draw a tree.

Then: go along the given branch.

Then: read the next branch.

Then: simplify.

Last: check.

Answer
P(2nd green | 1st yellow) = 23

Your turn

Try this one yourself, then check:

Show the answer
P(2nd green | 1st yellow) = 34

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