Teach Me Maths

Tree Diagrams

Question 11 mark

The probability that Kemi's bus is late on any day is 0.15

Here is a probability tree diagram for Monday and Tuesday. It is not complete.

Work out the probability that should go on the branch marked with a question mark.

Hint

The two branches from the same point add up to 1.

Worked solution

The probability that Kemi's bus is late on any day is 0.15

Here is a probability tree diagram for Monday and Tuesday. It is not complete.

Work out the probability that should go on the branch marked with a question mark.

  1. On Monday the bus is either late or on time.
  2. P(on time) =1-0.15=0.85.

Question 21 mark

Lucy takes at random a counter from bag A. She then takes at random a counter from bag B.

Here is a probability tree diagram. It is not complete.

Work out the probability that should go on the branch marked with a question mark.

Hint

From each point, the probabilities on the branches add up to 1.

Worked solution

Lucy takes at random a counter from bag A. She then takes at random a counter from bag B.

Here is a probability tree diagram. It is not complete.

Work out the probability that should go on the branch marked with a question mark.

  1. For bag B, the counter is red or blue.
  2. P(blue) =1-\frac{3}{5}=\frac{2}{5}.

Question 32 marks

Rosa plays two games.

The probability that she wins game 1 is 0.7

The probability that she wins game 2 is 0.4

Here is a probability tree diagram. It is not complete.

(a)

Work out the probability that should go on the branch labelled a.

1 mark

(b)

Work out the probability that should go on each branch labelled b.

1 mark

Hint

From each point, the probabilities on the two branches add up to 1.

Worked solution

Rosa plays two games.

The probability that she wins game 1 is 0.7

The probability that she wins game 2 is 0.4

Here is a probability tree diagram. It is not complete.

Part (a)

Work out the probability that should go on the branch labelled a.

  1. Rosa either wins or loses game 1.
  2. 1-0.7=0.3.

Part (b)

Work out the probability that should go on each branch labelled b.

  1. Rosa either wins or loses game 2.
  2. 1-0.4=0.6.

Question 42 marks

Dan takes two penalties. The probability tree diagram shows the probabilities that he will score or miss each penalty.

Work out the probability that Dan will score both penalties.

Hint

Multiply the probabilities along the path for score then score.

Worked solution

Dan takes two penalties. The probability tree diagram shows the probabilities that he will score or miss each penalty.

Work out the probability that Dan will score both penalties.

  1. Follow score on the first penalty, then score on the second.
  2. Multiply along the branches: 0.8\times0.8=0.64.

Question 52 marks

Ali drives through two sets of traffic lights. The probability tree diagram shows the probabilities that each set of lights will be red when he gets there.

Work out the probability that neither set of lights will be red.

Hint

Neither red means not red at the first lights and not red at the second lights.

Worked solution

Ali drives through two sets of traffic lights. The probability tree diagram shows the probabilities that each set of lights will be red when he gets there.

Work out the probability that neither set of lights will be red.

  1. Follow not red, then not red.
  2. Multiply along the branches: \frac{2}{3}\times\frac{3}{4}=\frac{6}{12}.
  3. \frac{6}{12}=\frac{1}{2}.

Question 61 mark

Maya takes two tests.

The probability that she passes the first test is 0.9

The probability that she passes the second test is 0.6

The two results do not affect each other.

Which calculation gives the probability that Maya passes both tests?

Hint

For one outcome and then another, follow one path along the tree.

Worked solution

Maya takes two tests.

The probability that she passes the first test is 0.9

The probability that she passes the second test is 0.6

The two results do not affect each other.

Which calculation gives the probability that Maya passes both tests?

  1. 0.9+0.6
  2. 0.9\times0.6
  3. 0.9\times0.4
  4. 0.1\times0.4
  1. Passing both means pass, then pass: one path along the tree.
  2. Multiply the probabilities along the path: 0.9\times0.6.

Question 72 marks

Jack plays one game of darts and then one game of pool.

The probability tree diagram shows the probabilities that he will win or lose each game.

Work out the probability that Jack wins both games.

Hint

Follow win, then win, and multiply.

Worked solution

Jack plays one game of darts and then one game of pool.

The probability tree diagram shows the probabilities that he will win or lose each game.

Work out the probability that Jack wins both games.

  1. Follow win at darts (0.75), then win at pool (0.6).
  2. 0.75\times0.6=0.45.

Question 83 marks

Kai takes a theory test and a hazard test.

The probability that Kai passes the theory test is 0.7

The probability that Kai passes the hazard test is 0.8

The two results do not affect each other.

(a)

Work out the probability that Kai passes both tests.

1 mark

(b)

Work out the probability that Kai does not pass both tests. (He fails one test or fails both tests.)

2 marks

Hint

Passing both tests is one path. Everything else is not passing both.

Worked solution

Kai takes a theory test and a hazard test.

The probability that Kai passes the theory test is 0.7

The probability that Kai passes the hazard test is 0.8

The two results do not affect each other.

Part (a)

Work out the probability that Kai passes both tests.

  1. Multiply along the path pass, pass: 0.7\times0.8=0.56.

Part (b)

Work out the probability that Kai does not pass both tests. (He fails one test or fails both tests.)

  1. Either Kai passes both tests or he does not.
  2. 1-0.56=0.44.

Question 93 marks

Zara enters a maths quiz and a science quiz.

The probability tree diagram shows the probabilities that she will win or not win each quiz.

Work out the probability that Zara wins exactly one of the two quizzes.

Hint

Exactly one win can happen in two ways. Work out each path, then add.

Worked solution

Zara enters a maths quiz and a science quiz.

The probability tree diagram shows the probabilities that she will win or not win each quiz.

Work out the probability that Zara wins exactly one of the two quizzes.

  1. Win maths, not win science: 0.6\times0.7=0.42.
  2. Not win maths, win science: 0.4\times0.3=0.12.
  3. Add the two paths: 0.42+0.12=0.54.

Question 104 marks

A bag contains 3 red counters and 5 blue counters.

Ella takes at random a counter from the bag. She puts the counter back in the bag.

She then takes at random a second counter from the bag.

Here is a probability tree diagram. It is not complete.

(a)

Work out the probability that should go on the branch labelled a.

1 mark

(b)

Work out the probability that should go on the branch labelled b.

1 mark

(c)

Work out the probability that both counters are red.

2 marks

Hint

The counter is put back, so the bag is the same for the second counter.

Worked solution

A bag contains 3 red counters and 5 blue counters.

Ella takes at random a counter from the bag. She puts the counter back in the bag.

She then takes at random a second counter from the bag.

Here is a probability tree diagram. It is not complete.

Part (a)

Work out the probability that should go on the branch labelled a.

  1. 3 of the 8 counters are red.
  2. P(red) =\frac{3}{8}.

Part (b)

Work out the probability that should go on the branch labelled b.

  1. The first counter is put back, so there are still 5 blue counters out of 8.
  2. P(blue) =\frac{5}{8}.

Part (c)

Work out the probability that both counters are red.

  1. Follow red, then red.
  2. \frac{3}{8}\times\frac{3}{8}=\frac{9}{64}.

Question 113 marks

Luca takes two shots at a target.

The probability that he hits the target with his first shot is 0.4

The probability that he hits the target with his second shot is 0.35

The two shots do not affect each other.

(a)

Work out the probability that Luca misses the target with both shots.

2 marks

(b)

Work out the probability that Luca hits the target with at least one of his shots.

1 mark

Hint

Missing means not hitting, so first work out the probability of a miss for each shot.

Worked solution

Luca takes two shots at a target.

The probability that he hits the target with his first shot is 0.4

The probability that he hits the target with his second shot is 0.35

The two shots do not affect each other.

Part (a)

Work out the probability that Luca misses the target with both shots.

  1. P(miss first) =1-0.4=0.6. P(miss second) =1-0.35=0.65.
  2. Multiply along the path miss, miss: 0.6\times0.65=0.39.

Part (b)

Work out the probability that Luca hits the target with at least one of his shots.

  1. Hitting with at least one shot is every outcome except missing with both shots.
  2. 1-0.39=0.61.

Question 123 marks

Noah spins a biased spinner twice.

The probability tree diagram shows the probabilities that the spinner lands on red.

(a)

Work out the probability that the spinner lands on red both times.

1 mark

(b)

Work out the probability that the spinner lands on red exactly once.

2 marks

Hint

Write down every path that gives the outcome, then add the paths.

Worked solution

Noah spins a biased spinner twice.

The probability tree diagram shows the probabilities that the spinner lands on red.

Part (a)

Work out the probability that the spinner lands on red both times.

  1. Follow red, then red: 0.7\times0.7=0.49.

Part (b)

Work out the probability that the spinner lands on red exactly once.

  1. Red then not red: 0.7\times0.3=0.21.
  2. Not red then red: 0.3\times0.7=0.21.
  3. 0.21+0.21=0.42.

Question 131 mark

Ravi plays two games. The result of the first game does not affect the probability of winning the second game.

The probability that he wins the first game is 0.4

The probability that he wins the second game is 0.5

Ravi says, "The probability that I win both games is 0.4+0.5=0.9"

What mistake has Ravi made?

Hint

Think about what happens to the chance of winning when he has to win two games instead of one.

Worked solution

Ravi plays two games. The result of the first game does not affect the probability of winning the second game.

The probability that he wins the first game is 0.4

The probability that he wins the second game is 0.5

Ravi says, "The probability that I win both games is 0.4+0.5=0.9"

What mistake has Ravi made?

  1. He should multiply the probabilities, not add them
  2. He should use 1-0.4 and 1-0.5
  3. He should add 0.6 and 0.5
  4. He has not made a mistake
  1. Winning both games means following the win-then-win path on a tree.
  2. The results are independent: after a win in the first game, the chance of winning the second is still 0.5. So 0.5 of the 0.4 first-game win chance leads to two wins.
  3. Multiply along this path: 0.4\times0.5=0.2. Ravi should multiply the probabilities, not add them.
  4. Check: winning both cannot be more likely than winning the first game alone, whose probability is 0.4. So 0.9 cannot be right.

Question 144 marks

Ella has a bag of 9 sweets. 4 of the sweets are mints and the other 5 are toffees.

She also has a tin of 7 biscuits. 2 of the biscuits are chocolate and the other 5 are plain.

Ella takes at random one sweet from the bag and one biscuit from the tin.

Here is a probability tree diagram. It is not complete.

(a)

Work out the probability that should go on the branch labelled a.

1 mark

(b)

Work out the probability that should go on the branch labelled b.

1 mark

(c)

Work out the probability that Ella takes a toffee and a plain biscuit.

2 marks

Hint

The sweet Ella takes does not change what is in the tin of biscuits.

Worked solution

Ella has a bag of 9 sweets. 4 of the sweets are mints and the other 5 are toffees.

She also has a tin of 7 biscuits. 2 of the biscuits are chocolate and the other 5 are plain.

Ella takes at random one sweet from the bag and one biscuit from the tin.

Here is a probability tree diagram. It is not complete.

Part (a)

Work out the probability that should go on the branch labelled a.

  1. 5 of the 9 sweets are toffees.
  2. P(toffee) =\frac{5}{9}.

Part (b)

Work out the probability that should go on the branch labelled b.

  1. The biscuit comes from the tin, so the sweet makes no difference.
  2. 2 of the 7 biscuits are chocolate: P(chocolate) =\frac{2}{7}.

Part (c)

Work out the probability that Ella takes a toffee and a plain biscuit.

  1. Follow toffee (\frac{5}{9}), then plain (\frac{5}{7}).
  2. \frac{5}{9}\times\frac{5}{7}=\frac{25}{63}.

Question 152 marks

A box contains 10 pens. 7 of the pens are black and 3 of the pens are blue.

Tom takes at random a pen from the box. He does not put it back.

He then takes at random a second pen from the box.

The probability tree diagram shows this information.

Work out the probability that both pens are blue.

Hint

Follow blue, then blue, and multiply the fractions.

Worked solution

A box contains 10 pens. 7 of the pens are black and 3 of the pens are blue.

Tom takes at random a pen from the box. He does not put it back.

He then takes at random a second pen from the box.

The probability tree diagram shows this information.

Work out the probability that both pens are blue.

  1. Follow blue (\frac{3}{10}), then blue (\frac{2}{9}).
  2. \frac{3}{10}\times\frac{2}{9}=\frac{6}{90}.
  3. \frac{6}{90}=\frac{1}{15}.

Question 16Challenge3 marks

Spinner A is a fair 5-sided spinner numbered 1, 2, 3, 4 and 5

Spinner B is a fair 3-sided spinner numbered 1, 2 and 3

Mia spins each spinner once.

The probability tree diagram shows the probabilities that each spinner lands on an even number.

Work out the probability that one spinner lands on an even number and the other spinner lands on an odd number.

Hint

This can happen in two ways: A even and B odd, or A odd and B even.

Worked solution

Spinner A is a fair 5-sided spinner numbered 1, 2, 3, 4 and 5

Spinner B is a fair 3-sided spinner numbered 1, 2 and 3

Mia spins each spinner once.

The probability tree diagram shows the probabilities that each spinner lands on an even number.

Work out the probability that one spinner lands on an even number and the other spinner lands on an odd number.

  1. A even, B odd: \frac{2}{5}\times\frac{2}{3}=\frac{4}{15}.
  2. A odd, B even: \frac{3}{5}\times\frac{1}{3}=\frac{3}{15}.
  3. Add the two paths: \frac{4}{15}+\frac{3}{15}=\frac{7}{15}.

Question 17Challenge4 marks

The probability that Omar scores a goal in a hockey match is 0.4

Omar is going to play in two hockey matches. Whether he scores in one match does not affect the other match.

Omar has drawn this probability tree diagram for the two matches. One of the probabilities on it is wrong.

(a)

Which probability is wrong?

1 mark

(b)

What should the wrong probability be?

1 mark

(c)

Work out the probability that Omar does not score a goal in either match.

2 marks

Hint

The two branches from the same point must add up to 1.

Worked solution

The probability that Omar scores a goal in a hockey match is 0.4

Omar is going to play in two hockey matches. Whether he scores in one match does not affect the other match.

Omar has drawn this probability tree diagram for the two matches. One of the probabilities on it is wrong.

Part (a)

Which probability is wrong?

  1. The 0.4 for goal in the 1st match
  2. The 0.6 for no goal in the 1st match
  3. The 0.4 for goal in the 2nd match, after no goal in the 1st match
  4. The 0.5 for no goal in the 2nd match, after no goal in the 1st match
  1. P(no goal) =1-0.4=0.6 in each match.
  2. After no goal in the 1st match, the branches show 0.4 and 0.5, which add up to 0.9, not 1.
  3. So the 0.5 is wrong.

Part (b)

What should the wrong probability be?

  1. 1-0.4=0.6.

Part (c)

Work out the probability that Omar does not score a goal in either match.

  1. Follow no goal, then no goal, using the correct probabilities.
  2. 0.6\times0.6=0.36.

Question 18Challenge4 marks

Rovers are playing two football matches.

The probability that Rovers win the first match is 0.7

The probability that Rovers win the second match is 0.45

The result of the first match does not affect the second match.

Here is a probability tree diagram. It is not complete.

(a)

Work out the probability that should go on the branch labelled x.

1 mark

(b)

Work out the probability that Rovers do not win either match.

2 marks

(c)

Which is more likely?

1 mark

Hint

Not winning either match means not win, then not win: one path along the tree.

Worked solution

Rovers are playing two football matches.

The probability that Rovers win the first match is 0.7

The probability that Rovers win the second match is 0.45

The result of the first match does not affect the second match.

Here is a probability tree diagram. It is not complete.

Part (a)

Work out the probability that should go on the branch labelled x.

  1. P(win second match) =0.45 whatever happens in the first match.
  2. x=1-0.45=0.55.

Part (b)

Work out the probability that Rovers do not win either match.

  1. Follow not win (0.3), then not win (0.55).
  2. 0.3\times0.55=0.165.

Part (c)

Which is more likely?

  1. Rovers win both matches
  2. Rovers do not win either match
  3. They are equally likely
  1. P(win both) =0.7\times0.45=0.315.
  2. 0.315 is more than 0.165, so winning both matches is more likely.

Question 19Challenge5 marks

Kim takes a written test and a practical test.

The probability that she passes the written test is 0.8

The probability that she passes the practical test is 0.55

The two results do not affect each other.

(a)

Work out the probability that Kim passes exactly one of the two tests.

3 marks

(b)

Kim says, "I am more likely to pass both tests than to pass exactly one test." Is Kim right?

2 marks

Hint

Draw a tree diagram. Passing exactly one test can happen in two ways.

Worked solution

Kim takes a written test and a practical test.

The probability that she passes the written test is 0.8

The probability that she passes the practical test is 0.55

The two results do not affect each other.

Part (a)

Work out the probability that Kim passes exactly one of the two tests.

  1. Pass written, fail practical: 0.8\times0.45=0.36.
  2. Fail written, pass practical: 0.2\times0.55=0.11.
  3. 0.36+0.11=0.47.

Part (b)

Kim says, "I am more likely to pass both tests than to pass exactly one test." Is Kim right?

  1. Yes, because 0.8 and 0.55 are both more than 0.5
  2. No, because P(passes both) =0.44, which is less than P(passes exactly one)
  3. Yes, because P(passes both) =0.8+0.55, which is more than P(passes exactly one)
  4. No, because P(passes both) =0.2\times0.45=0.09
  1. P(passes both) =0.8\times0.55=0.44.
  2. 0.44 is less than 0.47, so Kim is not right.

Question 20Challenge5 marks

A drawer contains 4 black socks and 2 white socks.

Jo takes at random a sock from the drawer. She writes down its colour and puts it back in the drawer.

Jo then takes at random a second sock from the drawer.

Here is a probability tree diagram. It is not complete.

(a)

Work out the probability that should go on the branch labelled a.

1 mark

(b)

Work out the probability that should go on the branch labelled b.

1 mark

(c)

Work out the probability that Jo takes two socks of the same colour.

3 marks

Hint

The first sock is put back, so the drawer is the same for the second sock. For the same colour, add the paths black, black and white, white.

Worked solution

A drawer contains 4 black socks and 2 white socks.

Jo takes at random a sock from the drawer. She writes down its colour and puts it back in the drawer.

Jo then takes at random a second sock from the drawer.

Here is a probability tree diagram. It is not complete.

Part (a)

Work out the probability that should go on the branch labelled a.

  1. The first sock is put back, so there are still 4 black socks out of 6.
  2. \frac{4}{6} (or \frac{2}{3}).

Part (b)

Work out the probability that should go on the branch labelled b.

  1. The first sock is put back, so there are still 2 white socks out of 6.
  2. \frac{2}{6} (or \frac{1}{3}).

Part (c)

Work out the probability that Jo takes two socks of the same colour.

  1. Black, black: \frac{4}{6}\times\frac{4}{6}=\frac{16}{36}.
  2. White, white: \frac{2}{6}\times\frac{2}{6}=\frac{4}{36}.
  3. Add: \frac{16}{36}+\frac{4}{36}=\frac{20}{36}=\frac{5}{9}.

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