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Either paper 3 marks
A bag contains 2 blue tiles and 7 grey tiles.
A tile is taken from the bag at random, its colour is written down, and it is put back in the bag.
A second tile is then taken from the bag at random.
The tree diagram is not complete.
(a)State the probability that the second tile taken is grey.
(b)Find the probability that both tiles taken are blue.
Show answer
(a) \(\frac{7}{9}\)
(b) \(\frac{4}{81}\)The first tile is put back before the second one is taken, so the bag is exactly as it started: 9 tiles, 7 of them grey.
\(P(\text{second is grey}) = \frac{ 7 }{ 9 }\)
Because the tile goes back, the second pair of branches carries the same two probabilities as the first pair.
Follow the top path through the tree and multiply along it.
\(P(\text{first is blue}) = \frac{ 2 }{ 9 }\), and because the tile is put back, \(P(\text{second is blue}) = \frac{ 2 }{ 9 }\) as well.
\(\frac{ 2 }{ 9 } \times \frac{ 2 }{ 9 } = \frac{ 4 }{ 81 }\)
- (b) M1multiplying along the top path: \(\frac{ 2 }{ 9 } \times \frac{ 2 }{ 9 }\)
- (b) A1correct probability \(\frac{ 4 }{ 81 }\)
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Either paper 3 marks
The probability that the train is late on Tuesday is \(0.4\).
The probability that it is late on Wednesday is \(0.1\).
The tree diagram shows this information. One probability is missing.
(a)State the probability that should be written on the branch marked with a question mark.
(b)Find the probability that the train is on time on both days.
Show answer
(a) \(0.6\)
(b) \(0.54\)The two branches for Tuesday are the only two things that can happen, so their probabilities add up to 1.
\(1 - 0.4 = 0.6\)
On time on Wednesday has probability \(1 - 0.1 = 0.9\).
On time on both days is the bottom path through the tree, so multiply along it:
\(0.6 \times 0.9 = 0.54\)
- (b) M1multiplying the on-time probabilities \(0.6 \times 0.9\)
- (b) A1correct probability \(0.54\)
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Either paper 3 marks
Owen and Leila each take a piano exam.
The probability that Owen passes is \(0.7\).
The probability that Leila passes is \(0.3\).
The two results are independent. The tree diagram shows Owen's result, then Leila's result.
(a)Complete the tree diagram.
(b)Work out the probability that Owen fails and Leila passes.
Show answer
(a) Owen fails \(0.3\). Leila under Owen passing: fails \(0.7\). Leila under Owen failing: passes \(0.3\), fails \(0.7\).
(b) \(0.09\)Each pair of branches from one point adds up to 1.
\(P(\text{Owen fails}) = 1 - 0.7 = 0.3\)
\(P(\text{Leila fails}) = 1 - 0.3 = 0.7\)
The two results are independent, so Leila's branches are the same whatever Owen does: \(0.3\) for a pass and \(0.7\) for a fail, on both of Owen's branches.
Owen fails and Leila passes, which is one path through the tree. Multiply along it:
\(0.3 \times 0.3 = 0.09\)
- (a) B1Owen fails \(0.3\) and Leila fails \(0.7\) under a pass
- (a) B1Leila branches repeated under a fail: \(0.3\) and \(0.7\)
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Either paper 3 marks
A fair, ordinary dice is rolled and a counter is taken at random from a bag.
The tree diagram shows the probabilities.
(a)How do the probabilities show that all the counters in the bag are white, blue or green?
(b)State the probability that the counter is white or green.
(c)State the probability that the dice lands on a number greater than 4 and the counter is white.
Show answer
(a) The three probabilities from each counter node add up to 1: \(\frac{ 2 }{ 20 } + \frac{ 9 }{ 20 } + \frac{ 9 }{ 20 } = 1\), so no other colour is possible.
(b) \(\frac{11}{20}\)
(c) \(\frac{1}{30}\)Each counter node has three branches, and the probabilities on them must cover every possible outcome.
\(\frac{ 2 }{ 20 } + \frac{ 9 }{ 20 } + \frac{ 9 }{ 20 } = \frac{ 20 }{ 20 } = 1\)
The probabilities add up to 1, so there is no probability left for any other colour: every counter is white, blue or green.
The counter is white or green, so add the two branch probabilities:
\(\frac{ 2 }{ 20 } + \frac{ 9 }{ 20 } = \frac{11}{20}\)
Follow the path More than 4, then white, and multiply the probabilities along it:
\(\frac{1}{ 3 } \times \frac{ 2 }{ 20 } = \frac{1}{30}\)
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Either paper 4 marks
Leila plays one game of darts and one game of snooker.
The probability that she wins the game of darts is \(0.8\).
The probability that she wins the game of snooker is \(0.2\).
The two results are independent.
(a)Find the probability that Leila wins both games.
(b)Find the probability that Leila wins neither game.
Show answer
(a) \(0.16\)
(b) \(0.16\)Winning both games is the top path through the tree, so multiply along it.
\(0.8 \times 0.2 = 0.16\)
\(P(\text{loses darts}) = 1 - 0.8 = 0.2\)
\(P(\text{loses snooker}) = 1 - 0.2 = 0.8\)
Losing both is the bottom path through the tree:
\(0.2 \times 0.8 = 0.16\)
- (a) M1multiplies the two win probabilities \(0.8 \times 0.2\)
- (a) A1P(both) \(0.16\)
- (b) M1multiplies the two losing probabilities \(0.2 \times 0.8\)
- (b) A1P(neither) \(0.16\)
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Either paper 3 marks
On any Monday, the probability that Amara is late for school is \(\frac{ 5 }{ 9 }\).
On the same Monday, the probability that Ben is late for school is \(\frac{ 5 }{ 6 }\).
Whether one of them is late does not affect the other.
Work out the probability that exactly one of Amara and Ben is late.
Show answer
\(\frac{25}{54}\)Exactly one late means one of them is late and the other is not, so there are two different paths through the tree and both of them count.
Amara late, Ben on time:
\(\frac{ 5 }{ 9 } \times \frac{ 1 }{ 6 } = \frac{ 5 }{ 54 }\)
Amara on time, Ben late:
\(\frac{ 4 }{ 9 } \times \frac{ 5 }{ 6 } = \frac{ 20 }{ 54 }\)
Different paths through a tree are added:
\(\frac{ 5 }{ 54 } + \frac{ 20 }{ 54 } = \frac{ 25 }{ 54 } = \frac{25}{54}\)
- M1one path found: \(\frac{ 5 }{ 9 } \times \frac{ 1 }{ 6 } = \frac{ 5 }{ 54 }\)
- M1other path found: \(\frac{ 4 }{ 9 } \times \frac{ 5 }{ 6 } = \frac{ 20 }{ 54 }\)
- A1added: \(\frac{ 25 }{ 54 }\), i.e. \(\frac{25}{54}\)
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Either paper 5 marks
Isla takes three shots at a netball goal.
The probability that she scores with any one shot is \(0.4\), and the shots are independent of each other.
(a)Work out the probability that Isla scores with all three shots.
(b)Work out the probability that Isla scores with at least one of the three shots.
Show answer
(a) \(0.064\)
(b) \(0.784\)Scoring three times is the top path through the tree, so multiply along all three branches:
\(0.4 \times 0.4 \times 0.4 = 0.064\)
"At least one" covers seven of the eight paths, so work out the one path it does not cover and take it away from 1.
\(P(\text{misses with one shot}) = 1 - 0.4 = 0.6\)
\(P(\text{misses all three}) = 0.6 \times 0.6 \times 0.6 = 0.216\)
\(P(\text{at least one}) = 1 - 0.216 = 0.784\)
- (a) M1multiplies along the three branches: \(0.4 \times 0.4 \times 0.4\)
- (a) A1probability \(0.064\)
- (b) M1misses all three: \(0.6 \times 0.6 \times 0.6\)
- (b) M1subtracts from 1: \(1 - 0.216\)
- (b) A1probability \(0.784\)
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Either paper 4 marks
There are 40 girls and 80 boys in a choir.
If a girl is chosen at random, the probability that she plays a musical instrument is 0.3.
If a boy is chosen at random, the probability that he plays a musical instrument is 0.49.
One of the 120 people is chosen at random.
Show that the probability that this person plays a musical instrument is more than 0.4.
Show answer
\(\frac{ 51.2 }{ 120 } \approx 0.427\), which is more than 0.4The two groups are different sizes, so each probability is weighted by its group's share of the 120 people.
Share of girls: \(\frac{ 40 }{ 120 }\). Share of boys: \(\frac{ 80 }{ 120 }\).
The probability is
\(\frac{ 40 }{ 120 } \times 0.3 + \frac{ 80 }{ 120 } \times 0.49\)
\(= \frac{ 12 }{ 120 } + \frac{ 39.2 }{ 120 }\)
\(= \frac{ 51.2 }{ 120 } \approx 0.427\)
\(0.427\) is more than \(0.4\), so the probability is more than 0.4.
- B1group shares \(\frac{ 40 }{ 120 }\) and \(\frac{ 80 }{ 120 }\)
- M1the two products \(\frac{ 12 }{ 120 }\) and \(\frac{ 39.2 }{ 120 }\)
- M1their sum \(\frac{ 51.2 }{ 120 } \approx 0.427\)
- A1more than 0.4, comparing 0.427 with 0.4
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Either paper 3 marks
Leila plays two games of Scrabble against Owen.
In each game, Leila could win, draw or lose.
In each game, the probability that Leila wins is \(0.22\).
In each game, the probability that Leila draws is \(0.15\).
Calculate the probability that Leila wins exactly one of the two games.
Show answer
\(0.3432\)Not winning covers a draw and a loss together, so the tree only needs two branches: win and not win. The probability of a draw is not needed.
\(P(\text{not win}) = 1 - 0.22 = 0.78\)
Winning exactly one game means win then not win, or not win then win.
Win, then not win:
\(0.22 \times 0.78 = 0.1716\)
Not win, then win:
\(0.78 \times 0.22 = 0.1716\)
Different paths are added:
\(0.1716 + 0.1716 = 0.3432\)
- M1P(not win) = 1 - \(0.22\) = \(0.78\)
- M1both paths \(0.22 \times 0.78 = 0.1716\)
- A1sum of the two paths \(0.3432\)
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Either paper 6 marks
Rosa plays a game with an ordinary, fair dice.
If she rolls a 1 or a 2 she wins.
If she rolls a 3 or a 4 she loses.
If she rolls any other number she rolls again.
When she has to roll again, she wins if she rolls a 6. Otherwise she loses.
Draw a tree diagram for the game and write the probability on each branch.
Is Rosa more likely to win or to lose?
You must work out the probability that Rosa wins.
Show answer
\(P(\text{win}) = \frac{7}{18}\), which is less than \(\frac{1}{2}\), so Rosa is more likely to loseThe dice has six faces. Rosa wins on 2 of them and loses on 2 of them, so rolls again on \(6 - 2 - 2 = 2\).
First roll: win \(\frac{ 2 }{6}\), lose \(\frac{ 2 }{6}\), roll again \(\frac{ 2 }{6}\).
Second roll: Rosa wins when the dice shows a 6, which is 1 of the six faces. The second roll wins with probability \(\frac{ 1 }{6}\) and loses with probability \(\frac{ 5 }{6}\), and these branches come only after roll again.
Rosa can win on the first roll, or by rolling again and then winning:
\(P(\text{win}) = \frac{ 2 }{6} + \frac{ 2 }{6} \times \frac{ 1 }{6} = \frac{1}{3} + \frac{1}{18}\)
\(P(\text{win}) = \frac{7}{18}\)
This is less than \(\frac{1}{2}\), so Rosa is more likely to lose.
- B1tree with first-roll branches \(\frac{ 2 }{6}\) win, \(\frac{ 2 }{6}\) lose and \(\frac{ 2 }{6}\) roll again
- B1second-roll branches \(\frac{ 1 }{6}\) win and \(\frac{ 5 }{6}\) lose, drawn after roll again only
- M1path through roll again to a win: \(\frac{ 2 }{6} \times \frac{ 1 }{6}\)
- M1adding the direct win \(\frac{ 2 }{6}\) to that path
- A1P(win) = \(\frac{7}{18}\)
- C1more likely to lose, comparing \(\frac{7}{18}\) with \(\frac{1}{2}\)