Fairness and Expected Outcomes: Practise Online

Either paper 3 marks

A fair spinner has 5 equal sections. 1 of the sections is brown.

The spinner is spun 85 times.

(a)
How many times would you expect the spinner to land on brown?

(b)
How many times would you expect the spinner not to land on brown?
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(a) \(17\)
(b) \(68\)
On each spin the probability of brown is \(\frac{ 1 }{ 5 }\).

\(85 \times \frac{ 1 }{ 5 } = 17\)

So expect about 17 landings on brown.
The other 4 sections are not brown, so

\(85 - 17 = 68\)
- (a) M1probability \(\frac{ 1 }{ 5 }\) multiplied by the number of spins, \(85 \times \frac{ 1 }{ 5 }\)
- (a) A117 landings
Either paper 3 marks

A fair spinner has 10 equal sections. 4 of the sections are orange.

Hafsa spins it 100 times.

(a)
Work out how many times Hafsa would expect the spinner to land on orange.

(b)
In fact the spinner landed on orange 39 times.

Explain why this does not mean the spinner is not fair.
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(a) \(40\)
(b) An expected number is only what happens on average — a fair spinner will not land on orange exactly 40 times every 100 spins, and 39 is close to 40.
The probability of orange on one spin is \(\frac{ 4 }{ 10 }\).

\(100 \times \frac{ 4 }{ 10 } = 40\)
The expected number says what would happen on average over many sets of 100 spins. Any one set will usually differ from it a little.

\(39\) is close to \(40\), which is exactly what a fair spinner would be expected to produce.
- (a) M1probability \(\frac{ 4 }{ 10 }\) multiplied by the number of spins, \(100 \times \frac{ 4 }{ 10 }\)
- (a) A140 times
Either paper 4 marks

A spinner has 17 equal sections.

3 of the sections are yellow, 7 are green and 7 are red.

(a)
Is the spinner equally likely to land on each of the three colours?

Give a reason for your answer.

(b)
Find the probability that the spinner lands on yellow.
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(a) No — 3 sections are yellow, 7 are green and 7 are red, so the three colours do not have the same number of sections.
(b) \(\frac{ 3 }{ 17 }\)
The sections are all the same size, so a colour is more likely the more sections it has.

3 sections are yellow, 7 are green and 7 are red, so the three colours do not have the same number of sections.

The answer is No.
\(3\) of the 17 sections are yellow, so the probability is

\(\frac{ 3 }{ 17 }\)
- (a) M13 sections are yellow, 7 are green and 7 are red
- (a) A1No — the three colours do not have the same number of sections
- (b) M1number of yellow sections \(3\) as the numerator, or the total \(17\) as the denominator
- (b) A1probability \(\frac{ 3 }{ 17 }\)
Either paper 3 marks

The probability of tails when a biased coin is thrown is 0.3.

The coin is thrown 1450 times.

Work out the expected number of heads.
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\(1015\)
The outcome asked about is the other one, so its probability is 1 minus the stated probability.

\(1 - 0.3 = 0.7\)

Expected number = probability \(\times\) number of trials.

\(0.7 \times 1450 = 1015\)
- M1probability of the other outcome \(1 - 0.3 = 0.7\)
- M1multiplies by the number of trials \(0.7 \times 1450\)
- A1expected number 1015
Either paper 4 marks

Here are two spinners. On each spinner every section is equally likely.

Spinner A has 10 equal sections. 4 of them are red.

Spinner B has 6 equal sections. 1 of them is red.

You win a prize if the spinner lands on red.

(a)
Which spinner gives the better chance of winning a prize? Show your working.

(b)
Aisha uses spinner A 60 times. How many prizes would you expect Aisha to win?
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(a) Spinner A
(b) \(24\)
Write both probabilities with the common denominator 30:

\(\frac{ 4 }{ 10 } = \frac{ 12 }{ 30 }, \qquad \frac{ 1 }{ 6 } = \frac{ 5 }{ 30 }\)

\(\frac{ 12 }{ 30 }\) is the larger probability, so spinner A gives the better chance.
On spinner A the probability of winning is \(\frac{ 4 }{ 10 }\).

\(60 \times \frac{ 4 }{ 10 } = 24\)

So expect about 24 prizes.
- (a) M1common denominator, \(\frac{ 4 }{ 10 } = \frac{ 12 }{ 30 }\) and \(\frac{ 1 }{ 6 } = \frac{ 5 }{ 30 }\)
- (a) A1Spinner A, since \(\frac{ 12 }{ 30 }\) is the larger probability
- (b) M1probability \(\frac{ 4 }{ 10 }\) multiplied by the number of spins, \(60 \times \frac{ 4 }{ 10 }\)
- (b) A124 prizes
Either paper 2 marks

A spinner has 8 equal sections. 3 of the sections are white.

The spinner is spun a number of times. It is expected to land on white 27 times.

Work out how many spins there were.
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\(72\)
On each turn the probability of white is \(\frac{ 3 }{ 8 }\).

The expected number is the number of spins multiplied by that probability, so the number of spins is the expected number divided by it:

\(27 \div \frac{ 3 }{ 8 } = 27 \times \frac{ 8 }{ 3 } = 72\)
- M1dividing the expected number by the probability, \(27 \div \frac{ 3 }{ 8 }\)
- A172 spins
Either paper 3 marks

A spinner has 4 equal sections, of which 3 are orange.

Aisha spins it 65 times.

(a)
Work out how many times Aisha would expect the spinner to land on orange.

(b)
Explain what your answer to part (a) means, given that the spinner cannot land on orange a fraction of a time.
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(a) \(48.75\)
(b) It is an average over many sets of 65 spins — any one set will give a whole number, usually 48 or 49.
The probability of orange on one spin is \(\frac{ 3 }{ 4 }\).

\(65 \times \frac{ 3 }{ 4 } = 48.75\)
An expected number is an average, not a prediction of one particular set of spins.

Over many sets of 65 spins the number of orange results would average \(48.75\). Any single set gives a whole number, most often 48 or 49.
- (a) M1\(65\) multiplied by the probability \(\frac{ 3 }{ 4 }\)
- (a) A1expected number \(48.75\)
Either paper 4 marks

In a game, a spinner with 10 equal sections is spun once. 3 of the sections are purple, 2 are orange and 5 are yellow.

Maya wins if the spinner lands on purple. Jonah wins if the spinner lands on orange or on yellow.

(a)
Is this game fair? Give a reason for your answer.

(b)
The spinner is spun 110 times.

Work out an estimate for the number of times Maya wins.
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(a) No — Maya wins on 3 of the 10 sections and Jonah on 2 + 5 = 7, which is not the same, so the game is not fair.
(b) \(33\)
A game is fair when both players are equally likely to win.

Maya wins on the 3 purple sections, so has probability \(\frac{ 3 }{ 10 }\).

Jonah wins on the orange and the yellow sections, \(2 + 5 = 7\) of them, so has probability \(\frac{ 7 }{ 10 }\).

These are not the same, so the game is not fair.
Maya wins with probability \(\frac{ 3 }{ 10 }\).

\(\frac{ 3 }{ 10 } \times 110 = 33\)
- (a) M1Maya wins with probability \(\frac{ 3 }{ 10 }\), Jonah with probability \(\frac{ 7 }{ 10 }\)
- (a) A1No, comparing 3 with 7
- (b) M1probability \(\frac{ 3 }{ 10 }\) multiplied by the number of spins, \(\frac{ 3 }{ 10 } \times 110\)
- (b) A133 wins
Either paper 4 marks

There are 1500 students at a college.

A student is chosen at random.

(a)
The probability that the student is taking a GCSE resit is \(0.06\).

How many of the students are taking a GCSE resit?

(b)
The probability that the student is studying maths is \(0.68\).

The probability that the student is studying physics is \(0.51\).

Show that some students are studying maths and physics.
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(a) \(90\)
(b) The probabilities add to \(1.19\), which is more than 1, so the two groups must overlap: some students are studying both maths and physics.
The probability is the fraction of the students taking a resit, so multiply it by the number of students:

\(1500 \times 0.06 = 90\)
Add the two probabilities:

\(0.68 + 0.51 = 1.19\)

If nobody studied both, the two groups would be separate and the probabilities could add to no more than 1. \(1.19\) is more than 1, so the groups must overlap.

So some students are studying maths and physics. (As numbers of students, \(1020 + 765 = 1785\), which is more than \(1500\), so at least \(285\) students are in both groups.)
- (a) M1multiplying the number of students by the probability, \(1500 \times 0.06\)
- (a) A190 students
- (b) M1adding the two probabilities, \(0.68 + 0.51 = 1.19\)
- (b) A1more than 1, so some students must be in both groups
Either paper 6 marks

A church fete has a bottle tombola.

It costs \(50\)p for one go.

The probability that a player draws a winning ticket is \(\frac{1}{ 10 }\).

A player who wins is given \(£1.00\).

On one day the tombola has \(100\) goes.

(a)
Find the total amount of money taken by the tombola that day.

(b)
Find the total amount the stall would expect to pay out in prizes.

(c)
Does the tombola expect to make a profit or a loss that day, and by how much?

You must show your working.
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(a) \(£50.00\)
(b) \(£10.00\)
(c) A profit of £40.00
Every go costs \(50\)p, and there were \(100\) goes.

\(100 \times 50 = 5000\) pence

\(5000\text{p} = £50.00\)
One go in \(10\) is expected to win, so out of \(100\) goes the tombola expects

\(100 \div 10 = 10\) winners

Each of them takes \(£1.00\).

\(10 \times £1.00 = £10.00\)
Compare the money in with the money out.

Money in: \(£50.00\). Money out: \(£10.00\).

\(£50.00 - £10.00 = £40.00\)

The larger of the two is the money in, so the tombola expects a profit of \(£40.00\).
- (a) M1number of goes multiplied by the cost of a go, \(100 \times 50 = 5000\) pence
- (a) A1takings of £50.00
- (b) M1expected number of winners, \(100 \div 10 = 10\), multiplied by the prize \(£1.00\)
- (b) A1prize money of £10.00
- (c) M1compares money in and out, \(£50.00 - £10.00 = £40.00\)
- (c) A1A profit of £40.00