一个袋子里装有 10 个红珠子和 15 个绿珠子，它们除颜色外完全相同。从袋中随机逐个摸出珠子，每次摸完后放回。

1. Find the probability of drawing a green bead on any single draw

Total number of beads $= 10 + 15 = 25$.
$P(G) = \frac{15}{25} = \frac{3}{5}$

2. Use the complement method to calculate the required probability

The event that the first red bead is obtained on or before the 5th draw is the complement of the event that the first 5 draws are all green. Since draws are with replacement, they are independent events.

$P(\text{first red on or before 5th draw}) = 1 - P(5\text{ consecutive green beads})$
$= 1 - \left(\frac{3}{5}\right)^5$
$= 1 - \frac{243}{3125}$
$= \frac{2882}{3125} = 0.92224$

1. Determine the probabilities of a single draw and define the events

The probability of drawing a red bead is $P(R) = \frac{10}{25} = \frac{2}{5}$.
The probability of drawing a green bead is $P(G) = \frac{15}{25} = \frac{3}{5}$.
Let $X$ be the draw number on which the first green bead is obtained.
The event "no green bead has been obtained in the first 5 draws" means the first 5 draws were all red, corresponding to $X > 5$.

2. Express the required probability using the conditional probability formula

We need to find $P(X = 8 \mid X > 5)$. Using the conditional probability formula:
$P(X = 8 \mid X > 5) = \frac{P(X = 8 \text{ and } X > 5)}{P(X > 5)}$
Since $X = 8$ naturally implies $X > 5$, the intersection is simply $X = 8$.
$P(X = 8) = (P(R))^7 \times P(G) = \left(\frac{2}{5}\right)^7 \left(\frac{3}{5}\right)$
$P(X > 5) = (P(R))^5 = \left(\frac{2}{5}\right)^5$

3. Evaluate the conditional probability

$P(X = 8 \mid X > 5) = \frac{\left(\frac{2}{5}\right)^7 \left(\frac{3}{5}\right)}{\left(\frac{2}{5}\right)^5} = \left(\frac{2}{5}\right)^2 \left(\frac{3}{5}\right)$
$P(X = 8 \mid X > 5) = \frac{4}{25} \times \frac{3}{5} = \frac{12}{125}$

(Alternatively, because the draws are independent with replacement, the outcome is equivalent to getting 2 red beads followed by 1 green bead from a fresh starting point: $\frac{2}{5} \times \frac{2}{5} \times \frac{3}{5} = \frac{12}{125}$.)

1. Determine the probabilities of drawing each colour of bead.

$\mathrm{P}(\text{Red}) = \frac{10}{25} = \frac{2}{5}$
$\mathrm{P}(\text{Green}) = \frac{15}{25} = \frac{3}{5}$

2. Formulate the probability for the two mutually exclusive scenarios that meet the condition.

For both colours to be obtained for the first time on exactly the $r$-th draw, either:

• The first $r-1$ draws are red and the $r$-th draw is green, OR
• The first $r-1$ draws are green and the $r$-th draw is red.

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

3. Manipulate the expression into the required format $a\left(b^{r-2}+c^{r-2}\right)$.

Extract a single power of $\frac{2}{5}$ and $\frac{3}{5}$ from the $(r-1)$ terms to leave $(r-2)$ in the exponent:
$\mathrm{P}(\text{exactly } r \text{ draws}) = \left(\frac{2}{5}\right)\left(\frac{2}{5}\right)^{r-2}\left(\frac{3}{5}\right) + \left(\frac{3}{5}\right)\left(\frac{3}{5}\right)^{r-2}\left(\frac{2}{5}\right)$
$= \frac{6}{25}\left(\frac{2}{5}\right)^{r-2} + \frac{6}{25}\left(\frac{3}{5}\right)^{r-2}$
$= \frac{6}{25}\left[ \left(\frac{2}{5}\right)^{r-2} + \left(\frac{3}{5}\right)^{r-2} \right]$

1. Interpret the condition for the event to occur on the 5th draw or later.

For the first beads of different colours to be obtained in 5 or more draws ($r \ge 5$), it implies that the first 4 draws must all be of the exact same colour.

$P(r \ge 5) = P(\text{First 4 draws are Red}) + P(\text{First 4 draws are Green})$

2. Calculate the probability of getting 4 consecutive red or 4 consecutive green beads.

Since the beads are replaced, the probability of drawing a red bead is $P(R) = \frac{10}{25} = \frac{2}{5}$ and a green bead is $P(G) = \frac{15}{25} = \frac{3}{5}$.

$P(\text{4 Reds}) = \left(\frac{2}{5}\right)^4 = \frac{16}{625}$
$P(\text{4 Greens}) = \left(\frac{3}{5}\right)^4 = \frac{81}{625}$

3. Sum the probabilities to find the final result.

$P(r \ge 5) = \frac{16}{625} + \frac{81}{625}$
$P(r \ge 5) = \frac{97}{625}$