A bag contains 10 red and 15 green beads which are indistinguishable apart from colour. Beads are drawn from the bag singly and at random, with replacement.

1. Determine the probabilities of drawing a single red or green bead.

Total number of beads $= 10 + 15 = 25$
$\text{P(Red)} = \frac{10}{25} = \frac{2}{5}$
$\text{P(Green)} = \frac{15}{25} = \frac{3}{5}$

2. Calculate the probability using the complement method.

The event that the first red bead is obtained on or before the 5th draw is the complement of the event that no red bead is obtained in the first 5 draws (i.e., all first 5 draws are green).
Since the draws are with replacement, they are independent.
$\text{P(First red on or before 5th draw)} = 1 - \text{P(5 consecutive Greens)}$
$= 1 - \left(\frac{3}{5}\right)^5$
$= 1 - \frac{243}{3125}$
$= \frac{2882}{3125}$ (or $0.92224$)

1. Find the probability of drawing each colour in a single draw.

$P(R) = \frac{10}{25} = \frac{2}{5}$
$P(G) = \frac{15}{25} = \frac{3}{5}$

2. Set up the conditional probability.

Let $A$ be the event of getting the first green bead on the 8th draw (i.e. 7 red beads followed by 1 green bead).
Let $B$ be the event of getting no green bead in the first 5 draws (i.e. 5 red beads).

Since draws are with replacement, they are independent.
$P(A) = \left(\frac{2}{5}\right)^7 \left(\frac{3}{5}\right)$
$P(B) = \left(\frac{2}{5}\right)^5$

Because event $A$ is a subset of event $B$ (getting the first green on the 8th draw strictly implies no green in the first 5), $P(A \cap B) = P(A)$.

By the conditional probability formula:
$P(A \mid B) = \frac{P(A \cap B)}{P(B)} = \frac{P(A)}{P(B)}$

3. Evaluate the conditional probability.

$P(A \mid B) = \frac{\left(\frac{2}{5}\right)^7 \left(\frac{3}{5}\right)}{\left(\frac{2}{5}\right)^5} = \left(\frac{2}{5}\right)^2 \left(\frac{3}{5}\right)$
$P(A \mid B) = \frac{4}{25} \times \frac{3}{5} = \frac{12}{125}$

1. Identify the two mutually exclusive sequences of outcomes for exactly $r$ draws and write down their probabilities.

Let the probability of drawing a red bead be $P(R) = \frac{10}{25} = \frac{2}{5}$ and a green bead be $P(G) = \frac{15}{25} = \frac{3}{5}$.

For both colours to be obtained on exactly the $r$-th draw, the outcomes must be either $r-1$ consecutive red beads followed by a green bead, or $r-1$ consecutive green beads followed by a red bead.

$P(\text{exactly } r \text{ draws}) = P(R^{r-1}G) + P(G^{r-1}R)$
$= \left(\frac{2}{5}\right)^{r-1}\left(\frac{3}{5}\right) + \left(\frac{3}{5}\right)^{r-1}\left(\frac{2}{5}\right)$

2. Factorise the expression to manipulate it into the specified algebraic form.

We need the powers inside the bracket to be $r-2$. We can extract one factor of $\frac{2}{5}$ from the first term's power and one factor of $\frac{3}{5}$ from the second term's power:

$P(\text{exactly } r \text{ draws}) = \left(\frac{2}{5}\right)\left(\frac{2}{5}\right)^{r-2}\left(\frac{3}{5}\right) + \left(\frac{3}{5}\right)\left(\frac{3}{5}\right)^{r-2}\left(\frac{2}{5}\right)$
$= \frac{6}{25}\left(\frac{2}{5}\right)^{r-2} + \frac{6}{25}\left(\frac{3}{5}\right)^{r-2}$
$= \frac{6}{25}\left( \left(\frac{2}{5}\right)^{r-2} + \left(\frac{3}{5}\right)^{r-2} \right)$

1. Translate the condition into the sequence of draws required

For the first different colour to be obtained on the 5th draw or later ($\ge 5$ draws), the first 4 draws must all be of the exact same colour. They must either be all red or all green.

2. Calculate the probability of obtaining all red or all green in the first 4 draws

Probability of drawing a red bead, $P(R) = \frac{10}{25} = \frac{2}{5}$
Probability of drawing a green bead, $P(G) = \frac{15}{25} = \frac{3}{5}$

Since the beads are drawn with replacement, the draws are independent.
$P(\text{first 4 draws are the same colour}) = P(R,R,R,R) + P(G,G,G,G)$

3. Evaluate the final probability

$P(\text{first 4 draws are the same}) = \left(\frac{2}{5}\right)^4 + \left(\frac{3}{5}\right)^4$
$= \frac{16}{625} + \frac{81}{625}$
$= \frac{97}{625}$