The first cube is not put back, so one green cube has already gone and there are 7 cubes left in the bag.

Of those 7, exactly 4 are green:

\(P(\text{second is green} \mid \text{first is green}) = \frac{ 4 }{ 7 }\)

Follow the top path through the tree and multiply along it.

\(P(\text{first is green}) = \frac{ 5 }{ 8 }\)

\(P(\text{second is green} \mid \text{first is green}) = \frac{ 4 }{ 7 }\) — from part (a)

\(\frac{ 5 }{ 8 } \times \frac{ 4 }{ 7 } = \frac{ 20 }{ 56 } = \frac{5}{14}\)

This person is one of those who play a musical instrument, so only the 16 students in that circle can be chosen — not all 30.

Of those 16, 3 are also in the overlap, so also are in the school choir:

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

The overlap is still 3, but now divide by all those who are in the school choir, which is 7, not 30.

\(\frac{ 3 }{ 7 }\)

The overlap is \(\frac{ 2 }{ 3 }\) of 21:

\(21 \div 3 \times 2 = 14\)

So Job only is \(21 - 14 = 7\).

The number of students who do not have a part-time job is \(61 - 21 = 40\).

40% of these play football, so Football only is

\(\frac{ 40 }{100} \times 40 = 16\)

The rest are in neither set:

\(40 - 16 = 24\)

Only the students in Football can be chosen, so the denominator is that whole circle:

\(14 + 16 = 30\)

Of these, 14 are in the overlap, so also in Job:

\(\frac{ 14 }{ 30 }\), which is \(\frac{ 7 }{ 15 }\) in its simplest form

The same colour happens two ways — both red, or both blue — so there are two paths through the tree to find and add.

Both red: \(\frac{ 4 }{ 7 } \times \frac{ 3 }{ 6 } = \frac{ 12 }{ 42 }\)

Both blue: \(\frac{ 3 }{ 7 } \times \frac{ 2 }{ 6 } = \frac{ 6 }{ 42 }\)

The two paths are different outcomes, so add them:

\(\frac{ 12 }{ 42 } + \frac{ 6 }{ 42 } = \frac{ 18 }{ 42 } = \frac{3}{7}\)

First fill in the second pair of branches. One sweet has gone, so there are 10 sweets left every time, and how many of each is left depends on the branch above.

Different flavours happens two ways: toffee then mint, or mint then toffee.

Toffee then mint: \(\frac{ 7 }{ 11 } \times \frac{ 4 }{ 10 } = \frac{ 28 }{ 110 }\)

Mint then toffee: \(\frac{ 4 }{ 11 } \times \frac{ 7 }{ 10 } = \frac{ 28 }{ 110 }\)

Add the two paths:

\(\frac{ 28 }{ 110 } + \frac{ 28 }{ 110 } = \frac{ 56 }{ 110 } = \frac{28}{55}\)

A quicker route is the complement: the two sweets are either the same or different, and P(same) is \(\frac{ 54 }{ 110 }\), so P(different) is \(1 - \frac{ 54 }{ 110 } = \frac{ 56 }{ 110 }\).

Read it straight off the tree: take the Rain branch, then the Late branch that leaves it.

\(P(\text{late} \mid \text{rain}) = \frac{ 3 }{ 7 }\)

This is the top path through the tree, so multiply along it:

\(\frac{ 5 }{ 6 } \times \frac{ 3 }{ 7 } = \frac{ 15 }{ 42 }\)

Callum can be late on a rainy morning or on a dry one, so two paths through the tree end in Late.

Rain and late: \(\frac{ 5 }{ 6 } \times \frac{ 3 }{ 7 } = \frac{ 15 }{ 42 }\)

No rain and late: \(\frac{ 1 }{ 6 } \times \frac{ 1 }{ 7 } = \frac{ 1 }{ 42 }\)

Add the two paths:

\(\frac{ 15 }{ 42 } + \frac{ 1 }{ 42 } = \frac{ 16 }{ 42 } = \frac{8}{21}\)

The two sweets are either a matching pair or a different pair. Finding "different" directly means adding six routes (each ordered pair of two different flavours), so it is quicker to find P(same) and take it from 1.

After the first sweet is taken there are 13 sweets left, so each second pick has denominator 13.

Both toffee: \(\frac{ 2 \times 1 }{ 14 \times 13 }\)

Both mint: \(\frac{ 7 \times 6 }{ 14 \times 13 }\)

Both fruit: \(\frac{ 5 \times 4 }{ 14 \times 13 }\)

Add them, using the common denominator \(14 \times 13 = 182\):

\(P(\text{same}) = \frac{ 2 + 42 + 20 }{ 182 } = \frac{ 64 }{ 182 }\)

\(P(\text{different}) = 1 - \frac{ 64 }{ 182 } = \frac{ 118 }{ 182 } = \frac{59}{91}\)

All three red is the top path through the tree. Each time a red cube is taken, both the number of red cubes and the total drop by one.

\(P(\text{first is red}) = \frac{ 4 }{ 10 }\)

\(P(\text{second is red}) = \frac{ 3 }{ 9 }\)

\(P(\text{third is red}) = \frac{ 2 }{ 8 }\)

Multiply along the path:

\(\frac{ 4 }{ 10 } \times \frac{ 3 }{ 9 } \times \frac{ 2 }{ 8 } = \frac{ 24 }{ 720 } = \frac{1}{30}\)

"At least one winning" covers three of the four paths, so it is quicker to work out the one path it does not cover — both blank — and take that away from 1.

Both blank: \(\frac{ 5 }{ 8 } \times \frac{ 4 }{ 7 } = \frac{ 20 }{ 56 }\)

\(P(\text{at least one winning}) = 1 - \frac{ 20 }{ 56 } = \frac{ 36 }{ 56 } = \frac{9}{14}\)

The first disc is white, so take it out of the bag. There are now 29 discs: 10 white, 8 yellow and 11 purple. The first disc is already known, so its probability is not multiplied in.

The second and third discs must be the other two colours, in either order.

Yellow then purple: \(\frac{ 8 }{ 29 } \times \frac{ 11 }{ 28 } = \frac{ 88 }{ 812 }\)

Purple then yellow: \(\frac{ 11 }{ 29 } \times \frac{ 8 }{ 28 } = \frac{ 88 }{ 812 }\)

Add the two orders:

\(\frac{ 88 }{ 812 } + \frac{ 88 }{ 812 } = \frac{ 176 }{ 812 } = \frac{44}{203}\)