Add 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: \(3 \times 9 + 4 = 31\)

The four probabilities must add up to 1:

\(x + 3x + 2x + 0.1 = 1\)

\(6x = 1 - 0.1 = 0.1\)

\(x = 0.1 \div 6 = 0.15\)

\(P(\text{ purple }) = 3x = 3 \times 0.15 = 0.45\)

\(400 \times 0.45 = 180\)

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

\(\frac{ 7 }{ 24 } \times \frac{ 7 }{ 24 } = \frac{ 49 }{ 576 } = \frac{49}{576}\)

There are \(n\) counters and 6 of them are red, so

\(P(\text{first is red}) = \frac{ 6 }{ n }\)

One red counter has now gone, leaving 5 red counters out of \(n - 1\), so

\(P(\text{second is red}) = \frac{ 5 }{n - 1}\)

Multiply the two probabilities:

\(\frac{ 6 }{ n } \times \frac{ 5 }{n - 1} = \frac{ 30 }{n^2 - n}\)

This is equal to \(\frac{1}{ 29 }\), so

\(29 \times 30 = n^2 - n\)

\(n^2 - n - 870 = 0\)

Factorise \(n^2 - n - 870 = 0\):

\((n - 30)(n + 29) = 0\)

so \(n = 30\) or \(n = -29\).

A bag cannot hold a negative number of counters, so \(n = 30\).

Check: there are 6 red and 24 blue, and \(\frac{ 6 }{ 30 } \times \frac{ 5 }{ 29 } = \frac{1}{ 29 }\)

Let \(r\) be the number of red counters.

\(P(\text{both red}) = \frac{r}{ 30 } \times \frac{r - 1}{ 29 } = \frac{r^2 - r}{ 870 }\)

So

\(\frac{r^2 - r}{ 870 } = \frac{3}{29}\)

Multiply both sides by \(870\):

\(r^2 - r = 90\)

\(r^2 - r - 90 = 0\)

\((r - 10)(r + 9) = 0\)

so \(r = 10\) or \(r = -9\). A number of counters cannot be negative, so there are 10 red counters.

\(P(\text{red}) = \frac{1}{3}\) at the start, so the counters can be grouped into 3s: let there be \(3k\) counters, of which \(1k\) are red.

After the 9 extra red counters go in there are \(1k + 9\) red out of \(3k + 9\), and that probability is \(\frac{1}{2}\):

\(\frac{ 1k + 9 }{ 3k + 9 } = \frac{ 1 }{ 2 }\)

Cross-multiply:

\(2(1k + 9) = 1(3k + 9)\)

Expanding and collecting the \(k\) terms gives

\(k = 9\)

\(k = 9\)

So the bag started with \(3 \times 9 = 27\) counters (9 red and 18 green).

Check: \(\frac{ 18 }{ 36 } = \frac{1}{2}\)

The ratio \(4 : 5\) means that for every \(4 + 5 = 9\) balls, \(4\) are green and \(5\) are yellow.

\(P(\text{ green }) = \frac{ 4 }{ 9 }\) and \(P(\text{ yellow }) = \frac{ 5 }{ 9 }\)

The ball is replaced each time, so the probabilities stay the same on every draw.

Every ball taken is either green or yellow, so "at least one of each colour" is everything except "all green" and "all yellow". Those two cannot happen together.

\(P(\text{all } n \text{ are green}) = \left(\frac{ 4 }{ 9 }\right)^n\) and \(P(\text{all } n \text{ are yellow}) = \left(\frac{ 5 }{ 9 }\right)^n\)

\(P(\text{at least one of each}) = 1 - \left(\frac{ 4 }{ 9 }\right)^n - \left(\frac{ 5 }{ 9 }\right)^n\)

The plant is known to be in circle \(B\), so only the plants in circle \(B\) count. The group to divide by is the overlap plus the part of \(B\) outside the other circle, not the whole collection.

Circle \(B\) holds \((2x + 2) + (2x - 6) = 4x - 4\) plants, and the overlap is \(2x + 2\), so

\(\frac{ 2x + 2 }{ 4x - 4 } = \frac{ 3 }{ 4 }\)

Cross-multiply:

\(8x + 8 = 12x - 12\)

Collect the \(x\) terms and the numbers:

\(4x = 20\)

\(x = 5\)

The overlap is \(12\), the conditioned circle's own region is \(4\), the other circle's own region is \(8\) and the outside is \(4\).

Add every region, including the outside:

\(12 + 4 + 8 + 4 = 28\)

The ratio \(x : y = 1 : 3\) gives \(y = 3x\).

Green counters: \(3x - x - 8 = 2x - 8\)

The first counter is one of \(3x\); the second is one of \(3x - 1\) left, so each colour has probability

\(\frac{ \text{count} }{ 3x } \times \frac{ \text{count} - 1 }{ 3x - 1 }\)

Pink: \(\frac{ x }{ 3x } \times \frac{ x - 1 }{ 3x - 1 } = \frac{ x(x-1) }{ 3x(3x-1) }\)

Blue: \(\frac{ 8 }{ 3x } \times \frac{ 7 }{ 3x - 1 } = \frac{ 8(7) }{ 3x(3x-1) }\)

Green: \(\frac{ 2x - 8 }{ 3x } \times \frac{ 2x - 9 }{ 3x - 1 } = \frac{ (2x-8)(2x-9) }{ 3x(3x-1) }\)

Add the three over the common denominator \(3x(3x-1)\) and expand each product:

\(x(x-1) = x^2 - x\)

\(8(7) = 56\)

\((2x-8)(2x-9) = 4x^2 - 34x + 72\)

Numerator: \(x^2 - x + 56 + 4x^2 - 34x + 72 = 5x^2 - 35x + 128\)

Denominator: \(3x(3x-1) = 9x^2 - 3x\)

\(P(\text{same colour}) = \frac{ 5x^2 - 35x + 128 }{ 9x^2 - 3x }\)

Let the bag have \(u\) red and \(g\) green counters, so \(u + g\) counters in all.

First probability:

\(\frac{ g }{ u + g } = \frac{ 2 }{ 5 }\)

\(5g = 2(u + g)\)

\(2u - 3g = 0\)

After the extra counters there are \(g + 6\) green counters out of \(u + g + 8\):

\(\frac{ g + 6 }{ u + g + 8 } = \frac{ 5 }{ 9 }\)

\(9(g + 6) = 5(u + g + 8)\)

\(9g + 54 = 5u + 5g + 40\)

\(5u - 4g = 14\)

To make the \(u\) terms match, multiply the first equation by 5 and the second by 2:

\(10u - 15g = 0\)

\(10u - 8g = 28\)

Subtract the first equation from the second:

\(7g = 28\)

Divide both sides by the coefficient of \(g\):

\(g = 4\)

So there were \(4\) green counters. Put this into \(2u = 3g\):

\(2u = 3 \times 4 = 12\)

\(u = 12 \div 2 = 6\)

There were 6 red counters and 4 green counters.

Check: \(\frac{ 4 }{ 10 } = \frac{2}{5}\) and \(\frac{ 10 }{ 18 } = \frac{5}{9}\).