Question 7 (15 marks) Use a SEPARATE writing booklet - HSC - SSCE Mathematics Extension 2 - Question 7 - 2008 - Paper 1

To find the probability that the three balls drawn are of different colors, we need to identify the favorable ways to select one ball from each color.

1. Number of ways to choose one ball from each of the three colors: $n_r \cdot n_w \cdot n_b$

2. Total ways to draw three balls from the urn: $C(n_r + n_w + n_b, 3)$

Therefore, the probability is: $P_2 = \frac{n_r \cdot n_w \cdot n_b}{C(n_r + n_w + n_b, 3)}$

To determine the probability that two balls are of one color and the third is different, we consider each case:

1. Probability of two red and one other: $C(n_r, 2) \cdot (n_w + n_b)$

2. Similarly for other color combinations: $C(n_w, 2) \cdot (n_r + n_b)$ $C(n_b, 2) \cdot (n_r + n_w)$

Thus, the overall probability will be: $P_3 = \frac{C(n_r, 2) \cdot (n_w + n_b) + C(n_w, 2) \cdot (n_r + n_b) + C(n_b, 2) \cdot (n_r + n_w)}{C(n_r + n_w + n_b, 3)}$

Assuming n is large, we can reasonably estimate the probabilities:

1. Each term can be simplified assuming each color has the same number of balls, that is, $n_r \approx n_w \approx n_b \approx n$.

Thus:

• $p_1 \approx \frac{n^3}{C(3n, 3)} = \frac{1}{3}$
• $p_2 \approx \frac{n^3}{C(3n, 3)} = \frac{1}{3}$
• $p_3 \approx \frac{3n^2(2n)}{C(3n, 3)} = \frac{2}{3}$

This leads us to conclude: $p_2 : p_1 : p_3 \approx 1 : 1 : 2$.

Using properties of angles in a circle:

1. Identify that $\angle ZTSP$ is subtended by arc $ZS$ and similarly for $LTPs$, which leads to: $\angle ZTSP = \angle LTPs$ via the property that angles subtended by the same arc are equal.

To derive the ratio using the earlier result:

1. From triangle proportionality:

   • The angles established give rise to the relationships between the sides based on the sine rule.

2. Therefore:

   • Rearrange the sine rule to find the relationship between lengths a, b, and c leads to: $\frac{1}{a} = \frac{1}{b} + \frac{1}{c}$.

To derive the required expression:

1. Take the derivative of the velocity equation: $\frac{dv}{dt} = -\alpha e^{-\alpha t}(b - v_0)$

2. Substitute and reformulate to arrive at: $\frac{dv}{dt} = \alpha(b - v)$.

The parameter b represents the maximum velocity that the boat can achieve in the absence of any restraint. It indicates the limiting speed the boat approaches as time progresses.

To find the distance travelled...

1. Integrate the velocity function: $x(t) = \int v(t) dt$

2. Then, derive the expression and rearranging leads us to find: $x = \frac{b}{\alpha ln\left(\frac{b - v_0}{b - v}\right) + v_0 - v}{\alpha}$.

Using our established velocity function and knowing that:

1. Substitute $v = \frac{b}{2}$ into the equation for x: $x = \frac{b}{\alpha ln\left(\frac{b - v_0}{b - \frac{b}{2}}\right) + v_0 - \frac{b}{2}}{\alpha}$

2. This will yield the distance travelled based on time to reach that velocity.