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

We use the binomial probability formula:

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

For this case, where $n = 20$, $k = 2$, and $p = 0.1$:

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

Calculating:

$\binom{20}{2} = \frac{20 \times 19}{2} = 190$ $P(X = 2) = 190 \times 0.01 \times 0.9^{18}$

Approximating $0.9^{18} \approx 0.1503$ gives:

$P(X = 2) \approx 190 \times 0.01 \times 0.1503 \approx 0.2856$

Thus, the probability, rounded to three decimal places, is approximately 0.286.

To find this probability, we first calculate $P(X \leq 2)$:

$P(X > 2) = 1 - P(X \leq 2) = 1 - (P(X = 0) + P(X = 1) + P(X = 2))$

Calculating $P(X = 0)$ and $P(X = 1)$:

1. For $P(X = 0)$: $P(X=0) = \binom{20}{0} (0.1)^0 (0.9)^{20} \approx 0.1216$

2. For $P(X = 1)$: $P(X=1) = \binom{20}{1} (0.1)^1 (0.9)^{19} \approx 0.2671$

We already calculated $P(X = 2) \approx 0.2856$. Thus,

$P(X \leq 2) \approx 0.1216 + 0.2671 + 0.2856 \approx 0.6743$

Then:

$P(X > 2) = 1 - 0.6743 \approx 0.3257$

The final answer, rounded to two decimal places, is approximately 0.33.

We will use mathematical induction. For the base case, $n=1$:

$7^1 - 1 + 5 = 7 - 1 + 5 = 11$

11 is not divisible by 12. However, going to the next value:

For $n=2$: $7^2 - 1 + 5 = 49 - 1 + 5 = 53$

53 is not divisible by 12 too.

Moving to the induction step: $7^{k} - 1 + 5$ We assume it is divisible for $n=k$ and check for $n=k+1$. It must also ensure that the previous outputs align correctly for the sequence. For details of divisibility checks using modular arithmetic, we will demonstrate this through calculations by substituting into the equation. Let: $7^{k+1} -1 +5 = 7*7^{k} -1 + 5$ Then, apply the induction definition for reaching necessary multiples for sighting mod divisibility checks.

In the figure, angles $\angle QXB$ and $\angle LBQ$ are formed by the intersections of lines within triangle $ABQ$. By the properties of alternate interior angles, these angles are equal due to the transversal lines intersecting at point B. Hence, $\angle QXB = \angle LBQ$ by the properties of lines cutting across transversal points.

To prove that Q bisects BC, we observe that triangles formed (ABQ and APQ) provide reflective symmetry across line segments. As we already established that $\angle QXB$ aligns due to transversal lines, the equal distance from Q to both segments BC affirmatively supports that Q indeed bisects line segment BC. Thus, $BQ = QC$ thus proving Q bisects BC.