An urn contains n red balls, n white balls and n blue balls - HSC - SSCE Mathematics Extension 2 - Question 7 - 2008 - Paper 1

For this part, we need to choose one ball from each color. The number of ways to do this is simply the product of the number of balls of each color:

$p_d = \frac{n \cdot n \cdot n}{C(3n, 3)} = \frac{n^3}{C(3n, 3)}$.

To solve for this probability, we can consider three situations: two red and one from the other colors, two white and one from the other colors, and two blue and one from the other colors. Each scenario yields:

$p_m = 3 \cdot \frac{C(n, 2)C(n, 1)}{C(3n, 3)} = 3 \cdot \frac{\frac{n(n-1)}{2} \cdot n}{C(3n, 3)} = \frac{3n^2(n-1)}{2C(3n, 3)}$.

As n becomes large, we can approximate the combinations:

$C(n, 3) \approx \frac{n^3}{6}$ and ( C(3n, 3) \approx \frac{(3n)^3}{6} = \frac{27n^3}{6} ).

Thus, we find the ratios:

1. For $p_r$: $p_r \approx \frac{3 \cdot \frac{n^3}{6}}{\frac{27n^3}{6}} = \frac{3}{27} = \frac{1}{9}$.

2. For $p_d$: $p_d \approx \frac{n^3}{\frac{27n^3}{6}} = \frac{6}{27} = \frac{2}{9}$.

3. For $p_m$: $p_m \approx \frac{3n^2(n-1) / 2}{\frac{27n^3}{6}} = \frac{9(n-1)}{27} = \frac{n-1}{3}$.

Thus, for large n, the ratios become: $p_r : p_d : p_m \approx 1 : 2 : 3.$.

Since the tangential segment PT is perpendicular to radius at point P, and line QT is a secant intersecting the circle, we can apply the tangent-secant theorem to demonstrate:

$∠LZTP = ∠LPTS.$

Using the angles derived from step (i), we can apply the Angle Bisector Theorem: $\frac{a}{b} = \frac{PT}{PS} \text{ and } \frac{c}{b} = \frac{RS}{PT}$. Hence, $\frac{1}{a} = \frac{1}{b} + \frac{1}{c}$.

To differentiate the velocity equation with respect to time, we have:

$v = b - (b - v_0)e^{-\alpha t}$.

Differentiating gives: $\frac{dv}{dt} = \alpha(b - v).$

The physical significance of b represents the maximum velocity the boat can achieve when the effect of the current is fully considered. This is the terminal velocity of the boat.

To find the distance travelled x, we integrate the velocity function over time:

$x = \int_0^t v dt$. Combining the variables yields: $x = \frac{b}{\alpha} \log_e \left( \frac{b - v_0}{b - v} \right) + \frac{v_0 - v}{\alpha}$.

Plugging in $v = \frac{b}{2}$ into the distance function:

1. Calculate the time when $v$ reaches rac{b}{2} using the initial velocity function, and then substitute that time back into the integrated distance formula.