Chapter 5 of 8

All 8 chapters
  1. 1Listing all the outcomes, readunderstood
  2. 2Relative frequency and expected results, readunderstood
  3. 3The OR rule and the AND rule, readunderstood
  4. 4Tree diagrams, readunderstood
  5. 5Picking without putting back, readunderstood
  6. 6Venn diagrams and sets, readunderstood
  7. 7Conditional probability, readunderstood
  8. 8Counting choices, readunderstood

M4.6.5 Chapter 5

Picking without putting back

A bag has 5 red and 3 blue counters. I take one, keep it, and take a second. What is the chance that both are red?

Your guess

Pick one before you read on. It is fine to be wrong.

Your guess

  • You picked A. Read on. At the end, the chapter comes back to your guess.

  • You picked B. Read on. At the end, the chapter comes back to your guess.

  • You picked C. Read on. At the end, the chapter comes back to your guess.

By the end you can

  • Draw a tree diagram where the first pick changes the second
  • Find the chance that two picks from a bag are both red

In chapter 4 each counter was put back, so the second pick had the same chances as the first. Now the counter is kept. This is picking without putting back.

The bag has changed, so the second pick has different chances. The two picks are dependent: the first changes the chance of the second.

Draw it first

How the bag changes

  1. Step 1: a bag labelled the start, with a key red: 5 and blue: 3. Step 2: a bag labelled after a red is kept, with a key red: 4 and blue: 3.red: 5blue: 3the start

    1 The bag at the start has 5 red and 3 blue counters, 8 in all.

  2. Step 1: a bag labelled the start, with a key red: 5 and blue: 3. Step 2: a bag labelled after a red is kept, with a key red: 4 and blue: 3.red: 4blue: 3after a red is kept

    2 A red counter is taken and kept. Now the bag has 4 red and 3 blue counters, 7 in all.

A bag labelled after a blue is kept, with a key red: 5 and blue: 2.red: 5blue: 2after a blue is kept
If a blue counter is taken instead, the bag has 5 red and 2 blue.

Read the chances for the second pick from the bag as it now is. After a red is kept: red is 4/7 and blue is 3/7. After a blue is kept: red is 5/7 and blue is 2/7. In each case the two add to 1.

Draw it first

Draw the tree for picking without putting back

  1. Step 1: the first pick, a branch 5/8 to Red and a branch 3/8 to Blue. Step 2: the second pick. After Red, 4/7 to Red and 3/7 to Blue. After Blue, 5/7 to Red and 2/7 to Blue. Step 3: the ends RR: 5/14, RB: 15/56, BR: 15/56 and BB: 3/28.Red5/8Blue3/8

    1 Draw the first pick: red 5/8 and blue 3/8.

  2. Step 1: the first pick, a branch 5/8 to Red and a branch 3/8 to Blue. Step 2: the second pick. After Red, 4/7 to Red and 3/7 to Blue. After Blue, 5/7 to Red and 2/7 to Blue. Step 3: the ends RR: 5/14, RB: 15/56, BR: 15/56 and BB: 3/28.Red5/8Red4/7Blue3/7Blue3/8Red5/7Blue2/7

    2 Draw the second pick from each end. Use the bag as it is now: after red, 4/7 and 3/7. After blue, 5/7 and 2/7.

  3. Step 1: the first pick, a branch 5/8 to Red and a branch 3/8 to Blue. Step 2: the second pick. After Red, 4/7 to Red and 3/7 to Blue. After Blue, 5/7 to Red and 2/7 to Blue. Step 3: the ends RR: 5/14, RB: 15/56, BR: 15/56 and BB: 3/28.Red5/8Red4/7Blue3/7Blue3/8Red5/7Blue2/7RR: 5/14RB: 15/56BR: 15/56BB: 3/28

    3 Multiply along each path to write the chance at each end.

Worked example

A bag has 5 red and 3 blue counters. Two counters are taken without putting back. What is the probability that both are red?

  1. The first counter is red with probability 5/8.
  2. Then 4 red and 3 blue are left, so the second is red with probability 4/7.
  3. P(both red) = 5/8 × 4/7 = 20/56.
  4. 20/56 = 5/14.

Answer: 5/14

A tree diagram with two stages. From the start, a branch 5/8 leads to Red and a branch 3/8 leads to Blue. After Red, a branch 4/7 leads to Red and a branch 3/7 leads to Blue. After Blue, a branch 5/7 leads to Red and a branch 2/7 leads to Blue. The ends are printed RR: 5/14, RB: 15/56, BR: 15/56 and BB: 3/28. The end RR is hatched.Red5/8Red4/7Blue3/7Blue3/8Red5/7Blue2/7RR: 5/14RB: 15/56BR: 15/56BB: 3/28
The tree for picking without putting back. The hatched end is both red.

Worked example

The same bag, two counters taken and kept. What is the probability of one counter of each colour?

  1. Red then blue: 5/8 × 3/7 = 15/56.
  2. Blue then red: 3/8 × 5/7 = 15/56.
  3. The two paths are different, so add: 15/56 + 15/56 = 30/56.
  4. 30/56 = 15/28.

Answer: 15/28

Check the tree: the four ends must add to 1. Over 56, they are 20 + 15 + 15 + 6 = 56, so 56/56 = 1. This is a good way to catch a slip.

Compare with putting the counter back. The chance of both red falls from 25/64 to 5/14, because taking a red leaves fewer reds in the bag. The chance of one of each rises from 15/32 to 15/28.

A table with the head event, put back, kept. Rows: both red, 25/64, 5/14; one each, 15/32, 15/28; both blue, 9/64, 3/28.eventput backkeptboth red25/645/14one each15/3215/28both blue9/643/28
The same bag, put back or kept. Keeping the first counter makes "same colour" less likely.

Try it

Pick two stones and keep them

You need: 5 white stones and 3 dark stones, a cloth bag or a cup, a pencil and paper.

  1. Put the 8 stones in the bag. Pick two without looking, and write the colours.
  2. Put both back, shake the bag, and do it again.
  3. Do this 40 times. Count the times both were white.
  4. The tree says 5/14 of the pairs, so you expect about 14 of 40.

What you will see: Your count is near 14. If you put the first stone back each time, you would see more pairs of white stones, about 16 in 40.

If you cannot try it, this is what happens.
A table with the head pairs, two white, estimate. Rows: 40, 15, 0.375.pairstwo whiteestimate40150.375

A wrong idea many people have

The second pick has the same chances as the first.

Why it feels right: The bag looks the same, and the same colours are in it.

The test that breaks it: Take a red counter and keep it. The bag now has 7 counters, not 8, and only 4 are red. The chance of red is 4/7, not 5/8.

What is true: When a counter is kept, change both numbers for the second pick. There is one fewer in all, and one fewer of the colour taken.

Key words

without putting back
taking a counter and keeping it out, so the next pick is from a smaller bag.
dependent
events where one happening changes the chance of the other.

Back to your guess

The answer is B: 5/14, because 5/8 × 4/7.

After a red counter is taken and kept, 7 counters are left, and only 4 of them are red. So the second pick has chance 4/7, not 5/8. The chance of both is 5/8 × 4/7 = 20/56 = 5/14.

Check yourself

Try each one before you look. A wrong answer tells you where to go next.

  1. A bag has 4 red and 2 blue counters. A red counter is taken and kept. What is the chance that the next counter is red?

    A bag labelled the bag, with a key red: 4 and blue: 2.red: 4blue: 2the bag
    Tap your answer.

    1. Not this one. You used the bag as it was at the start. A red counter has gone, so the numbers change. Go to M4.6.5. Try another.

    2. Yes, that is right. After a red is kept, 3 red and 2 blue are left. That is 5 counters, so the chance of red is 3/5.

    3. Not this one. You took 1 off the red counters but forgot that the bag has 1 fewer counter in all. Go to M4.6.5. Try another.

    See the answer

    3/5

    After a red is kept, 3 red and 2 blue are left. That is 5 counters, so the chance of red is 3/5.

    • If you chose A: You used the bag as it was at the start. A red counter has gone, so the numbers change. Go to M4.6.5.
    • If you chose C: You took 1 off the red counters but forgot that the bag has 1 fewer counter in all. Go to M4.6.5.

  2. A bag has 4 red and 2 blue counters. Two are taken without putting back. What is the probability that both are red?

    Tap your answer.

    1. Not this one. You used 4/6 twice, as if the counter were put back. After a red is kept, the bag has 3 red in 5. Go to M4.6.5. Try another.

    2. Not this one. That is the chance for the first pick only. The second counter must be red too. Go to M4.6.5. Try another.

    3. Yes, that is right. P(both red) = 4/6 × 3/5 = 12/30 = 2/5.

    See the answer

    2/5

    P(both red) = 4/6 × 3/5 = 12/30 = 2/5.

    • If you chose A: You used 4/6 twice, as if the counter were put back. After a red is kept, the bag has 3 red in 5. Go to M4.6.5.
    • If you chose B: That is the chance for the first pick only. The second counter must be red too. Go to M4.6.5.

  3. A bag has 6 red and 3 blue counters. Two are taken without putting back. The first is blue. What is the chance that the second is red?

    Tap your answer.

    1. Yes, that is right. The blue counter is kept. Now 6 red and 2 blue are left, 8 in all, so the chance of red is 6/8 = 3/4.

    2. Not this one. That is the chance for the first pick. A blue counter has gone, so there are 8 left. Go to M4.6.5. Try another.

    3. Not this one. You took a red counter away, but the counter taken was blue. All 6 red are still in the bag. Go to M4.6.5. Try another.

    See the answer

    3/4

    The blue counter is kept. Now 6 red and 2 blue are left, 8 in all, so the chance of red is 6/8 = 3/4.

    • If you chose B: That is the chance for the first pick. A blue counter has gone, so there are 8 left. Go to M4.6.5.
    • If you chose C: You took a red counter away, but the counter taken was blue. All 6 red are still in the bag. Go to M4.6.5.

  4. A tree for two picks has three ends with chances 12/42, 12/42 and 12/42. The fourth is missing. What is it?

    Tap your answer.

    1. Not this one. Check by adding: 12 + 12 + 12 + 9 = 45, which is more than 42. The ends must add to 42/42. Go to M4.6.5. Try another.

    2. Yes, that is right. The ends of a tree add to 1, which is 42/42. 12 + 12 + 12 = 36, so the missing end is 42 − 36 = 6, which is 6/42.

    3. Not this one. Check by adding: 12 + 12 + 12 + 4 = 40, which is less than 42. The ends must add to 42/42. Go to M4.6.5. Try another.

    See the answer

    6/42

    The ends of a tree add to 1, which is 42/42. 12 + 12 + 12 = 36, so the missing end is 42 − 36 = 6, which is 6/42.

    • If you chose A: Check by adding: 12 + 12 + 12 + 9 = 45, which is more than 42. The ends must add to 42/42. Go to M4.6.5.
    • If you chose C: Check by adding: 12 + 12 + 12 + 4 = 40, which is less than 42. The ends must add to 42/42. Go to M4.6.5.

  5. A bag has 6 red and 4 blue counters. Two are taken without putting back. What is the probability that both are blue?

    A bag labelled the bag, with a key red: 6 and blue: 4.red: 6blue: 4the bag
    Tap your answer.

    1. Not this one. You used 2/5 twice, as if the counter were put back. After a blue is kept, the bag has 3 blue in 9. Go to M4.6.5. Try another.

    2. Not this one. That is the chance for the first pick alone. The second counter must be blue too. Go to M4.6.5. Try another.

    3. Not this one. That is the chance of blue on the second pick only, 3/9. Multiply by the chance of blue on the first pick, 4/10. Go to M4.6.5. Try another.

    4. Yes, that is right. P(both blue) = 4/10 × 3/9 = 12/90 = 2/15.

    See the answer

    2/15

    P(both blue) = 4/10 × 3/9 = 12/90 = 2/15.

    • If you chose A: You used 2/5 twice, as if the counter were put back. After a blue is kept, the bag has 3 blue in 9. Go to M4.6.5.
    • If you chose B: That is the chance for the first pick alone. The second counter must be blue too. Go to M4.6.5.
    • If you chose C: That is the chance of blue on the second pick only, 3/9. Multiply by the chance of blue on the first pick, 4/10. Go to M4.6.5.

  6. From an earlier chapter

    A bag has 2 red and 1 blue counter. One is taken, put back, and a second is taken. What is the probability that both are red?

    See the answer

    4/9

    Putting back keeps the chances the same: P(red) = 2/3 each time. 2/3 × 2/3 = 4/9.

  7. A bag has 4 red and 3 blue counters. Two are taken without putting back. What is the probability that both are red?

    See the answer

    2/7

    P(both red) = 4/7 × 3/6 = 12/42 = 2/7.

  8. A bag has 3 red and 2 blue counters. Two are taken without putting back. What is the probability that the two are different colours?

    See the answer

    3/5

    Red then blue: 3/5 × 2/4 = 6/20. Blue then red: 2/5 × 3/4 = 6/20. Add: 12/20 = 3/5.

Keep it

Copy this into your exercise book.

Without putting back, the bag shrinks. Second pick: one fewer counter in all, and one fewer of the colour taken. Multiply along paths. All ends add to 1.

Draw: A tree for two picks from a bag of 5 red and 3 blue, with 5/8, 3/8 and the changed second branches.

Reading with a child? A short page for the adult: what this chapter teaches and three questions to ask.

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