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Either paper 5 marks
A bag contains red, blue and green counters.
The number of red counters is \(x + 4\).
The number of blue counters is \(3x - 3\).
The number of green counters is \(3x + 4\).
A counter is taken from the bag at random. The probability that it is blue is \(\frac{6}{17}\).
(a)Find the value of \(x\).
(b)Work out how many counters of each colour are in the bag.
Show answer
(a) \(9\)
(b) 13 red, 24 blue and 31 greenAdd the three expressions to get the total number of counters:
\(\left(x + 4\right) + \left(3x - 3\right) + \left(3x + 4\right) = 7x + 5\)
The probability that a counter is blue is the blue expression over this total:
\(\frac{ 3x - 3 }{ 7x + 5 } = \frac{ 6 }{ 17 }\)
Cross-multiply:
\(17\left(3x - 3\right) = 6\left(7x + 5\right)\)
Expand and collect the \(x\) terms on one side:
\(9x = 81\)
\(x = 9\)
Substitute \(x = 9\) into each expression:
Red: \(9 + 4 = 13\)
Blue: \(3 \times 9 - 3 = 24\)
Green: \(3 \times 9 + 4 = 31\)
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Either paper 4 marks
A spinner can land on blue, purple, orange or red.
The table shows the probability of each colour.
Colour Blue Purple Orange Red Probability \(x\) \(3x\) \(2x\) \(0.1\) (a)Find the value of \(x\).
(b)The spinner is spun 400 times.
Work out an estimate for the number of times it lands on purple.
Show answer
(a) \(0.15\)
(b) \(180\)The four probabilities must add up to 1:
\(x + 3x + 2x + 0.1 = 1\)
\(6x = 1 - 0.1 = 0.1\)
\(x = 0.1 \div 6 = 0.15\)
\(P(\text{ purple }) = 3x = 3 \times 0.15 = 0.45\)
\(400 \times 0.45 = 180\)
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Either paper 4 marks
A bag contains \(24\) chips. Some of them are blue and the rest are yellow.
Omar takes a chip at random from the bag, records its colour, and puts it back.
He then takes a second chip at random from the bag.
The probability that both chips are blue is \(\frac{ 289 }{ 576 }\).
(a)Calculate how many of the chips in the bag are blue.
(b)Calculate the probability that neither chip is blue.
Show answer
(a) \(17\ \text{chips}\)
(b) \(\frac{49}{576}\)The first chip goes back before the second is taken, so the bag is the same both times and the two draws do not affect each other.
Write \(r\) for the number of blue chips. Each draw has probability \(\frac{r}{ 24 }\) of being blue, so
\(\frac{r}{ 24 } \times \frac{r}{ 24 } = \frac{ 289 }{ 576 }\)
\(\frac{r^2}{ 576 } = \frac{ 289 }{ 576 }\)
The denominators already match, so \(r^2 = 289\) and
\(r = \sqrt{ 289 } = 17\)
There are \(17\) blue chips in the bag.
\(24 - 17 = 7\) of the chips are yellow.
The bag is unchanged between the draws, so
\(\frac{ 7 }{ 24 } \times \frac{ 7 }{ 24 } = \frac{ 49 }{ 576 } = \frac{49}{576}\)
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Either paper 4 marks
A bag contains only red counters and green counters.
A counter is taken from the bag at random. The probability that it is red is \(\frac{1}{3}\).
The counter is put back, and then 8 more red counters are put into the bag.
A counter is now taken from the bag at random. The probability that it is red is \(\frac{1}{2}\).
Calculate how many counters were in the bag to start with.
Show answer
\(24\)\(P(\text{red}) = \frac{1}{3}\) at the start, so the counters can be grouped into 3s: let there be \(3k\) counters, of which \(1k\) are red.
After the 8 extra red counters go in there are \(1k + 8\) red out of \(3k + 8\), and that probability is \(\frac{1}{2}\):
\(\frac{ 1k + 8 }{ 3k + 8 } = \frac{ 1 }{ 2 }\)
Cross-multiply:
\(2(1k + 8) = 1(3k + 8)\)
Expanding and collecting the \(k\) terms gives
\(k = 8\)
\(k = 8\)
So the bag started with \(3 \times 8 = 24\) counters (8 red and 16 green).
Check: \(\frac{ 16 }{ 32 } = \frac{1}{2}\)
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Either paper 2 marks
There are only blue buttons and red buttons in a tin.
The number of blue buttons to the number of red buttons is in the ratio \(3 : 5\).
Nadia takes at random one button from the tin. She records the colour of the button and then replaces it in the tin.
Nadia does this \(n\) times, where \(n \geq 2\).
Write down an expression, in terms of \(n\), for the probability that Nadia gets at least one blue button and at least one red button.
Show answer
\(1 - \left(\frac{ 3 }{ 8 }\right)^n - \left(\frac{ 5 }{ 8 }\right)^n\)The ratio \(3 : 5\) means that for every \(3 + 5 = 8\) buttons, \(3\) are blue and \(5\) are red.
\(P(\text{ blue }) = \frac{ 3 }{ 8 }\) and \(P(\text{ red }) = \frac{ 5 }{ 8 }\)
The button is replaced each time, so the probabilities stay the same on every draw.
Every button taken is either blue or red, so "at least one of each colour" is everything except "all blue" and "all red". Those two cannot happen together.
\(P(\text{all } n \text{ are blue}) = \left(\frac{ 3 }{ 8 }\right)^n\) and \(P(\text{all } n \text{ are red}) = \left(\frac{ 5 }{ 8 }\right)^n\)
\(P(\text{at least one of each}) = 1 - \left(\frac{ 3 }{ 8 }\right)^n - \left(\frac{ 5 }{ 8 }\right)^n\)
- B1single-draw probabilities \(\frac{ 3 }{ 8 }\) for blue and \(\frac{ 5 }{ 8 }\) for red
- B1one minus P(all blue) minus P(all red), each a power of \(n\)
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Either paper 5 marks
There are only \(u\) red counters and \(g\) green counters in a bag.
A counter is taken at random from the bag. The probability that the counter is green is \(\frac{3}{7}\).
The counter is put back in the bag. 4 more red counters and 5 more green counters are put in the bag.
A counter is taken at random from the bag. The probability that the counter is green is \(\frac{7}{15}\).
Find the number of red counters and the number of green counters that were in the bag originally.
Show answer
\(12 red counters and 9 green counters\)Let the bag have \(u\) red and \(g\) green counters, so \(u + g\) counters in all.
First probability:
\(\frac{ g }{ u + g } = \frac{ 3 }{ 7 }\)
\(7g = 3(u + g)\)
\(3u - 4g = 0\)
After the extra counters there are \(g + 5\) green counters out of \(u + g + 9\):
\(\frac{ g + 5 }{ u + g + 9 } = \frac{ 7 }{ 15 }\)
\(15(g + 5) = 7(u + g + 9)\)
\(15g + 75 = 7u + 7g + 63\)
\(7u - 8g = 12\)
To make the \(u\) terms match, multiply the first equation by 7 and the second by 3:
\(21u - 28g = 0\)
\(21u - 24g = 36\)
Subtract the first equation from the second:
\(4g = 36\)
Divide both sides by the coefficient of \(g\):
\(g = 9\)
So there were \(9\) green counters. Put this into \(3u = 4g\):
\(3u = 4 \times 9 = 36\)
\(u = 36 \div 3 = 12\)
There were 12 red counters and 9 green counters.
Check: \(\frac{ 9 }{ 21 } = \frac{3}{7}\) and \(\frac{ 14 }{ 30 } = \frac{7}{15}\).
- M1first equation \(\frac{ g }{ u + g } = \frac{ 3 }{ 7 }\)
- M1second equation \(\frac{ g + 5 }{ u + g + 9 } = \frac{ 7 }{ 15 }\)
- M1both equations cleared of fractions into linear form: \(3u - 4g = 0\) and \(7u - 8g = 12\)
- A19 green counters, after eliminating \(u\)
- A112 red counters, by substituting back
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Either paper 5 marks
A bag contains red, blue and green counters.
The number of red counters is \(x + 6\).
The number of blue counters is \(3x + 6\).
The number of green counters is \(3x + 6\).
A counter is taken from the bag at random. The probability that it is blue is \(\frac{9}{22}\).
(a)Calculate the value of \(x\).
(b)Calculate how many counters of each colour are in the bag.
Show answer
(a) \(10\)
(b) 16 red, 36 blue and 36 greenAdd the three expressions to get the total number of counters:
\(\left(x + 6\right) + \left(3x + 6\right) + \left(3x + 6\right) = 7x + 18\)
The probability that a counter is blue is the blue expression over this total:
\(\frac{ 3x + 6 }{ 7x + 18 } = \frac{ 9 }{ 22 }\)
Cross-multiply:
\(22\left(3x + 6\right) = 9\left(7x + 18\right)\)
Expand and collect the \(x\) terms on one side:
\(3x = 30\)
\(x = 10\)
Substitute \(x = 10\) into each expression:
Red: \(10 + 6 = 16\)
Blue: \(3 \times 10 + 6 = 36\)
Green: \(3 \times 10 + 6 = 36\)
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Either paper 4 marks
A spinner can land on red, blue, green or yellow.
The table gives the probability of each colour.
Colour Red Blue Green Yellow Probability \(x\) \(3x\) \(2x\) \(0.4\) (a)Calculate the value of \(x\).
(b)The spinner is spun 300 times.
Work out an estimate for the number of times it lands on blue.
Show answer
(a) \(0.1\)
(b) \(90\)The four probabilities must add up to 1:
\(x + 3x + 2x + 0.4 = 1\)
\(6x = 1 - 0.4 = 0.4\)
\(x = 0.4 \div 6 = 0.1\)
\(P(\text{ blue }) = 3x = 3 \times 0.1 = 0.3\)
\(300 \times 0.3 = 90\)
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Either paper 4 marks
A bag contains \(15\) cubes. Some of them are green and the rest are white.
Zainab takes a cube at random from the bag, records its colour, and puts it back.
She then takes a second cube at random from the bag.
The probability that both cubes are green is \(\frac{ 169 }{ 225 }\).
(a)Calculate how many of the cubes in the bag are green.
(b)Find the probability that neither cube is green.
Show answer
(a) \(13\ \text{cubes}\)
(b) \(\frac{4}{225}\)The first cube goes back before the second is taken, so the bag is the same both times and the two draws do not affect each other.
Write \(r\) for the number of green cubes. Each draw has probability \(\frac{r}{ 15 }\) of being green, so
\(\frac{r}{ 15 } \times \frac{r}{ 15 } = \frac{ 169 }{ 225 }\)
\(\frac{r^2}{ 225 } = \frac{ 169 }{ 225 }\)
The denominators already match, so \(r^2 = 169\) and
\(r = \sqrt{ 169 } = 13\)
There are \(13\) green cubes in the bag.
\(15 - 13 = 2\) of the cubes are white.
The bag is unchanged between the draws, so
\(\frac{ 2 }{ 15 } \times \frac{ 2 }{ 15 } = \frac{ 4 }{ 225 } = \frac{4}{225}\)
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Either paper 4 marks
A bag contains only red counters and green counters.
A counter is taken from the bag at random. The probability that it is red is \(\frac{1}{4}\).
The counter is put back, and then 6 more red counters are put into the bag.
A counter is now taken from the bag at random. The probability that it is red is \(\frac{2}{5}\).
Work out how many counters were in the bag to start with.
Show answer
\(24\)\(P(\text{red}) = \frac{1}{4}\) at the start, so the counters can be grouped into 4s: let there be \(4k\) counters, of which \(1k\) are red.
After the 6 extra red counters go in there are \(1k + 6\) red out of \(4k + 6\), and that probability is \(\frac{2}{5}\):
\(\frac{ 1k + 6 }{ 4k + 6 } = \frac{ 2 }{ 5 }\)
Cross-multiply:
\(5(1k + 6) = 2(4k + 6)\)
Expanding and collecting the \(k\) terms gives
\(3k = 18\)
\(k = 6\)
So the bag started with \(4 \times 6 = 24\) counters (6 red and 18 green).
Check: \(\frac{ 12 }{ 30 } = \frac{2}{5}\)