The spinner must land on one of the four colours, so the four probabilities add up to 1.

\(0.35 + 0.15 + 0.1 = 0.6\)

\(1 - 0.6 = 0.4\)

So the probability of yellow is 0.4.

The mode is the score with the tallest bar, which is 24.

The bar for 26 has height 7, so the score was 26 in 7 of the 60 games.

So the estimated probability is \(\frac{ 7 }{ 60 }\).

The bars for 24 and every higher score have heights \(16,\ 14,\ 7\), which add up to 37.

So the winning score was 24 or more in 37 of the 60 games.

So the estimated probability is \(\frac{ 37 }{ 60 }\).

The dice has 6 equally likely faces, and 1 of them shows a number less than 2.

\(P(\text{dice less than 2}) = \frac{ 1 }{ 6 }\)

On the spinner there are 9 equal sections. Counting the sections whose number is greater than 5 gives 1.

\(P(\text{spinner greater than 5}) = \frac{ 1 }{ 9 }\)

To compare, write both fractions with denominator 18:

\(\frac{ 1 }{ 6 } = \frac{ 3 }{ 18 }\) and \(\frac{ 1 }{ 9 } = \frac{ 2 }{ 18 }\)

\(\frac{ 3 }{ 18 } > \frac{ 2 }{ 18 }\), so Anya is more likely to succeed on the dice.

3 of the 12 counters are black, so the probability is

\(\frac{ 3 }{ 12 } = \frac{ 1 }{ 4 }\)

The counter is either black or it is not, so the two probabilities add up to 1.

\(1 - \frac{ 1 }{ 4 } = \frac{ 3 }{ 4 }\)

Counting instead gives the same answer: \(12 - 3 = 9\) counters are not black, and \(\frac{ 9 }{ 12 } = \frac{ 3 }{ 4 }\)

\(\frac{ 1 }{ 4 }\) of the counters are purple, and that is 5 counters.

So one \(\frac{1}{ 4 }\) of the bag is \(5 \div 1 = 5\) counters, and the whole bag is

\(4 \times 5 = 20\)

\(20 - 5 = 15\)

Add the three probabilities:

\(0.3 + 0.4 + 0.4 = 1.1\)

The spinner must land on one of the three, so the probabilities have to add to exactly 1. \(1.1\) is more than 1, so at least one of them is wrong.

\(1.1 - 1 = 0.1\)

A counter cannot be two colours at once, so the two probabilities simply add:

\(0.3 + 0.25 = 0.55\)

Either it is one of those two colours or it is not, and the two probabilities add to 1.

\(1 - 0.55 = 0.45\)

The three probabilities add up to 1.

\(0.35 + 0.35 = 0.7\)

\(1 - 0.7 = 0.3\)

There are 3 numbers greater than 2, out of 5 altogether.

\(\frac{ 3 }{ 5 } = 0.6\)

\(0.6\) (the spinner) compared with \(0.3\) (an orange counter):

\(0.6 > 0.3\), so the spinner is more likely.

The probability of B is \(\frac{ 1 }{ 6 }\), so out of 12 sections B gets

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

The probability of A is three times the probability of B, so A gets

\(3 \times 2 = 6\)

That leaves \(12 - 2 - 6 = 4\) sections for C and D, and their probabilities must be equal, so each gets

\(4 \div 2 = 2\)

So B: 2, C: 2, D: 2, A: 6 — the sections can be arranged in any order around the spinner.

Each counter node has three branches, and the probabilities on them must cover every possible outcome.

\(\frac{ 6 }{ 25 } + \frac{ 12 }{ 25 } + \frac{ 7 }{ 25 } = \frac{ 25 }{ 25 } = 1\)

The probabilities add up to 1, so there is no probability left for any other colour: every counter is red, blue or yellow.

The counter is red or yellow, so add the two branch probabilities:

\(\frac{ 6 }{ 25 } + \frac{ 7 }{ 25 } = \frac{13}{25}\)

Follow the path Even, then red, and multiply the probabilities along it:

\(\frac{1}{ 2 } \times \frac{ 6 }{ 25 } = \frac{3}{25}\)