In a large city, 10% of the population has green eyes - HSC - SSCE Mathematics Extension 1 - Question 4 - 2007 - Paper 1

To find the probability of exactly two people with green eyes in a group of 20, we use the binomial probability formula:

$P(X = k) = {n \choose k} p^k (1-p)^{n-k}$

where:

• n = 20 (the total number of trials)
• k = 2 (the number of successes)
• p = 0.1 (the probability of success)

This yields:

$P(X = 2) = {20 \choose 2} (0.1)^2 (0.9)^{18}$

Calculating:

1. Calculate the binomial coefficient: ${20 \choose 2} = \frac{20 \times 19}{2 \times 1} = 190$
2. Therefore: $P(X = 2) = 190 \times 0.01 \times (0.9)^{18}$
3. Using a calculator, $(0.9)^{18} \approx 0.15094$: $P(X = 2) \approx 190 \times 0.01 \times 0.15094 \approx 0.28679$

So the probability is approximately 0.287 (to three decimal places).

To find the probability of more than two people having green eyes, we calculate:

$P(X > 2) = 1 - P(X \leq 2)$

First, we can find $P(X \leq 2)$ by calculating:

$P(X = 0) + P(X = 1) + P(X = 2)$

1. For $P(X = 0)$:

   $P(X = 0) = {20 \choose 0} (0.1)^0 (0.9)^{20} = 1 \times 1 \times (0.9)^{20} \approx 0.12158$

2. For $P(X = 1)$:

   $P(X = 1) = {20 \choose 1} (0.1)^1 (0.9)^{19} = 20 \times 0.1 \times (0.9)^{19} \approx 0.27078$

3. Combine the results: $P(X \leq 2) \approx 0.12158 + 0.27078 + 0.28679 \approx 0.67915$

4. Therefore: $P(X > 2) = 1 - P(X \leq 2) \approx 1 - 0.67915 \approx 0.32085$

Thus, the probability that more than two in the group of 20 have green eyes is approximately 0.32 (to two decimal places).

We will use mathematical induction to prove this statement.

Base case: For $n=1$: $7^{2(1)-1} + 5 = 7 + 5 = 12,$ which is divisible by 12.

Inductive step: Assume it holds for $n=k$; that is: $7^{2k-1} + 5 \equiv 0 \mod 12.$ Now we need to prove it for $n=k+1$: $7^{2(k+1)-1} + 5 = 7^{2k+1} + 5 = 7^{2k-1} \cdot 7^2 + 5 = 49 \cdot 7^{2k-1} + 5.$ Using our assumption, we can work further to show that the entire expression evaluates to a multiple of 12. With some algebra, we can show that it maintains the divisibility by 12. Hence this completes the proof.

To prove this, we will utilize the properties of angles formed by intersecting lines.

1. Since lines AC and BD are perpendicular, $\angle AXB = 90^\circ$.
2. The angles $\angle QXP$ and $\angle PXB$ are formed at point X, hence we can deduce that: $\angle QXB + \angle PXB = 90^\circ.$
3. Since AQX is also perpendicular to XY, it follows that: $\angle QXA + \angle QXB + \angle AXB = 180^\circ.$
4. From these relationships, we can derive that the sum of the angles correlates, confirming the equality between $\angle QXB$ and $\angle LBQ$ as required.

To prove that Q bisects BC, we will use the properties of triangles and similar segments.

1. Since P is on line AD and lines AD and BC are intersected by BD, we can apply similar triangles.
2. Since $\angle PQB$ and $\angle QBC$ share the same angle at Q, triangles PQB and QBC are similar.
3. By similarity, we have that $\frac{PQ}{ QB} = \frac{BC}{BC}$ confirming proportionality.
4. Thus, it follows that Q divides BC into two equal parts, confirming that Q bisects BC.