Each of the 8 numbers on the spinner can come up with a head or with a tail. Listing them systematically gives H1, H2, ... up to H8, and then T1, T2, ... up to T8.

\(2 \times 8 = 16\) outcomes.

The numbers greater than 1 are 2 up to 8, which is 7 numbers.

Each one gives one outcome with a head, so 7 of the 16 outcomes win.

\(P = \frac{ 7 }{ 16 }\)

As a fraction in its simplest form this is \(\frac{7}{16}\).

The more of the spinner a section takes up, the more likely the spinner is to land on it.

Section E is the largest, so the spinner is most likely to land on section E.

Sections A and D are the same size and no other two sections match, so the spinner is equally likely to land on sections A and D.

Each outcome is one section from each spinner, so a section appears many times in the list. Write each different one once.

The letters that appear are S, U, P, R, which makes 4 sections.

The numbers that appear are 1, 2, 3, which makes 3 sections.

Check: \(4 \times 3 = 12\), the number of outcomes in the list.

The grid has \(6 \times 6 = 36\) cells and each one is equally likely.

Counting the cells that show 4 gives 3 of them.

\(P(\text{total} = 4) = \frac{ 3 }{36}\)

As a fraction in its simplest form this is \(\frac{1}{12}\).

The totals greater than 9 are the ones in the bottom right corner of the grid. Counting them gives 6 cells out of 36.

\(P(\text{total} > 9) = \frac{ 6 }{36}\)

As a fraction in its simplest form this is \(\frac{1}{6}\).

The bottom row is for a score of 8 on spinner A, so add 8 to each of the spinner B numbers in turn:

\(8 + 6 = 14\), \(8 + 7 = 15\), \(8 + 8 = 16\), \(8 + 9 = 17\)

The completed grid has \(4 \times 4 = 16\) equally likely cells.

2 of them show a total of 16.

\(P(\text{total} = 16) = \frac{ 2 }{16}\)

As a fraction in its simplest form this is \(\frac{1}{8}\).

There are 36 equally likely cells in the grid.

A difference of 2 appears 8 times, half of them above the diagonal of zeros and half below it.

\(P(\text{difference} = 2) = \frac{ 8 }{36}\)

As a fraction in its simplest form this is \(\frac{2}{9}\).

Count the cells showing a difference bigger than 2. They sit in the two corners furthest from the diagonal, and there are 12 of them.

\(P(\text{difference} > 2) = \frac{ 12 }{36}\)

As a fraction in its simplest form this is \(\frac{1}{3}\).

The table has \(4\) rows and \(3\) columns, and every cell is one outcome.

\(4 \times 3 = 12\)

Count the even numbers in the table. There are \(6\) of them, out of \(12\) cells altogether.

\(P = \frac{ 6 }{12} = \frac{1}{2}\)

Go through the table and count the cells holding a number bigger than \(5\). There are \(7\) of them.

\(P = \frac{ 7 }{12} = \frac{7}{12}\)

For each cell compare the number on spinner A with the number on spinner B. If they are the same the score is 0, otherwise the score is the lower of the two numbers.

Row 5: 5 and 1 are different, so take the lower, score 1. 5 and 4 are different, so take the lower, score 4. 5 and 1 are different, so take the lower, score 1. 5 and 5 are the same, score 0.

Row 4: 4 and 1 are different, so take the lower, score 1. 4 and 4 are the same, score 0. 4 and 1 are different, so take the lower, score 1. 4 and 5 are different, so take the lower, score 4.

Row 6: 6 and 1 are different, so take the lower, score 1. 6 and 4 are different, so take the lower, score 4. 6 and 1 are different, so take the lower, score 1. 6 and 5 are different, so take the lower, score 5.

The completed table has \(3 \times 4 = 12\) equally likely cells, and a number on two sections of spinner B fills two columns.

The score 0 is even, so it is not counted as odd. There are \(7\) odd scores in the table.

\(P = \frac{ 7 }{12} = \frac{7}{12}\)

Use the one score shown in each of the columns p and q, with the number on spinner C for its row.

Column p, row 10: the score is 8. The score is not 0, so the numbers are different and the score is the lower one. It is below the number on C, so it is the number on D.

\(p = 8\)

Column q, row 8: the score is 7. The score is not 0, so the numbers are different and the score is the lower one. It is below the number on C, so it is the number on D.

\(q = 7\)

The top of the table is the first four positive odd numbers, so the missing number there is \(1\).

The side of the table is the first three positive odd numbers, so the missing number there is \(3\).

Each score is the number at the side multiplied by the number at the top. Multiplying each matching pair gives the three missing scores, in the order they are read along the rows: \(3 \times 1 = 3\), \(5 \times 3 = 15\) and \(5 \times 7 = 35\).

This is the completed table. Each score that is a square number is in bold.

×	1	3	5	7
1	1	3	5	7
3	3	9	15	21
5	5	15	25	35

The number of scores in bold is \(3\). The table has \(3 \times 4 = 12\) equally likely outcomes, so \(P = \frac{ 3 }{12}\) The highest common factor of \(3\) and \(12\) is \(3\). Dividing the top and the bottom by it gives the simplest form. \(P = \frac{1}{4}\)

Each of the 8 numbers on the spinner can come up with a head or with a tail. Listing them systematically gives H1, H2, ... up to H8, and then T1, T2, ... up to T8.

\(2 \times 8 = 16\) outcomes.

The numbers greater than 5 are 6 up to 8, which is 3 numbers.

Each one gives one outcome with a head, so 3 of the 16 outcomes win.

\(P = \frac{ 3 }{ 16 }\)

As a fraction in its simplest form this is \(\frac{3}{16}\).

The more of the spinner a section takes up, the more likely the spinner is to land on it.

Section B is the largest, so the spinner is most likely to land on section B.

Sections A and E are the same size and no other two sections match, so the spinner is equally likely to land on sections A and E.

Each outcome is one section from each spinner, so a section appears many times in the list. Write each different one once.

The letters that appear are V, Q, S, U, which makes 4 sections.

The numbers that appear are 1, 2, 3, which makes 3 sections.

Check: \(4 \times 3 = 12\), the number of outcomes in the list.