Steply

Probability with and

Probability · 20 problems, easy to harder

▶ Watch the first one solved

How we do it

  1. Independent or not
  2. Draw a tree
  3. Multiply along the branches
  4. Simplify
  5. Check

The steps for A bag has 6 red and 4 blue marbles, two are picked without…. Every solution names its steps, so you always know what comes next.

Worked examples

Flip a coin and roll a die, P(heads and 6) Easy

The question

Question: A coin is flipped and a die is rolled.

What is the probability of getting heads and a 6?

1 Independent or not

The coin does not change the die.

1st → does not change the 2nd

We call them independent.

independent:
the 1st does not change the 2nd

The die still has 6 sides, whatever the coin shows.

2 Draw a tree

We want heads, then a 6.

That is this path, in red.

3 Multiply along the branches

And means both happen.

We multiply along the path: 1/2, times 1/6.

P(heads and 6) = 12 × 16
and: multiply along the branches

Careful!

And means multiply, not add.

Careful: not 12 + 16 ✗

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

P(heads and 6) = 1 × 12 × 6

1 times 1 is 1.

2 times 6 is 12.

So it is 1/12.

P(heads and 6) = 112

4 Check

Let's check the tree.

Each end: multiply along its branches.

We do not cancel, so every bottom is 12.

All the ends together: 1 plus 5 plus 1 plus 5 is 12.

12/12 is 1.

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

112 + 512 + 112 + 512 = 1212 = 1

Let's look back.

First: independent or not.

Then: draw a tree.

Then: multiply along the branches.

Last: check.

Answer
P(heads and 6) = 112

▶ Watch it solved

A bag has 2 red and 3 blue marbles, two are picked with replacement… Medium

The question

Question: A bag has 2 red and 3 blue marbles.

Two marbles are picked, one after the other.

The first one is put back.

What is the probability the first is red and the second is blue?

1 Independent or not

The first marble is put back.

1st → does not change the 2nd

We call them independent.

independent:
the 1st does not change the 2nd

So the bag is the same for the second pick: 5 marbles.

2 Draw a tree

We want red, then blue.

That is this path, in red.

3 Multiply along the branches

And means both happen.

We multiply along the path: 2/5, times 3/5.

P(red then blue) = 25 × 35
and: multiply along the branches

Careful!

The marble is put back, so the second pick is just like the first.

Careful: put back: the same fractions again

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

P(red then blue) = 2 × 35 × 5

2 times 3 is 6.

5 times 5 is 25.

So it is 6/25.

P(red then blue) = 625

4 Check

Let's check the tree.

Each end: multiply along its branches.

We do not cancel, so every bottom is 25.

All the ends together: 4 plus 6 plus 6 plus 9 is 25.

25/25 is 1.

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

425 + 625 + 625 + 925 = 2525 = 1

Let's look back.

First: independent or not.

Then: draw a tree.

Then: multiply along the branches.

Last: check.

Answer
P(red then blue) = 625

▶ Watch it solved

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

The question

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

Two marbles are picked, one after the other.

The first one is not put back.

What is the probability both are blue?

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 blue.

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 blue) = 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 blue) = 4 × 310 × 9

4 times 3 is 12.

10 times 9 is 90.

So it is 12/90.

P(both blue) = 1290

4 Simplify

Both 12 and 90 can be divided by 6.

That makes it simpler.

P(both blue) = 12 ÷ 690 ÷ 6

So it is 2/15.

P(both blue) = 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: 30 plus 24 plus 24 plus 12 is 90.

90/90 is 1.

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

3090 + 2490 + 2490 + 1290 = 9090 = 1

Let's look back.

First: independent or not.

Then: draw a tree.

Then: multiply along the branches.

Then: simplify.

Last: check.

Answer
P(both blue) = 215

▶ Watch it solved

Practice

Try each one, then open the answer. Stuck? Open the problem to watch every step.

Easy

  1. flip a coin and roll a die, P(heads and 6)
    Answer
    P(heads and 6) = 112
  2. flip a coin and roll a die, P(tails and even)
    Answer
    P(tails and even) = 14
  3. roll a die twice, P(two 6s)
    Answer
    P(6 then 6) = 136
  4. flip a coin and roll a die, P(heads and more than 4)
    Answer
    P(heads and more than 4) = 16
  5. roll a die twice, P(6 then 1)
    Answer
    P(6 then 1) = 136
  6. a bag has 3 red and 5 blue marbles, two are picked with replacement, P(both red)
    Answer
    P(both red) = 964

Medium

  1. a bag has 2 red and 3 blue marbles, two are picked with replacement, P(red then blue)
    Answer
    P(red then blue) = 625
  2. a bag has 4 red and 4 blue marbles, two are picked with replacement, P(both blue)
    Answer
    P(both blue) = 14
  3. a bag has 3 red and 5 blue marbles, two are picked without replacement, P(both red)
    Answer
    P(both red) = 328
  4. a bag has 4 red and 6 blue marbles, two are picked without replacement, P(both red)
    Answer
    P(both red) = 215
  5. roll a die twice, P(even then odd)
    Answer
    P(even then odd) = 14
  6. a bag has 5 red and 3 blue marbles, two are picked without replacement, P(red then blue)
    Answer
    P(red then blue) = 1556
  7. flip a coin and roll a die, P(tails and less than 3)
    Answer
    P(tails and less than 3) = 16
  8. roll a die twice, P(odd then 6)
    Answer
    P(odd then 6) = 112

Harder

  1. a bag has 6 red and 4 blue marbles, two are picked without replacement, P(both blue)
    Answer
    P(both blue) = 215
  2. a bag has 2 green and 5 yellow marbles, two are picked without replacement, P(both green)
    Answer
    P(both green) = 121
  3. a bag has 3 red and 7 blue marbles, two are picked without replacement, P(blue then red)
    Answer
    P(blue then red) = 730
  4. roll a die twice, P(more than 4 then less than 3)
    Answer
    P(more than 4 then less than 3) = 19
  5. a bag has 5 white and 5 black marbles, two are picked without replacement, P(both black)
    Answer
    P(both black) = 29
  6. a bag has 4 red and 5 blue marbles, two are picked with replacement, P(blue then red)
    Answer
    P(blue then red) = 2081