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

To calculate the probability of exactly 2 out of 20 people having green eyes, we use the binomial probability formula:

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

Where:

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

Now substituting the values, we have:

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

Calculating this:

• $\binom{20}{2} = \frac{20 \times 19}{2} = 190$
• $0.1^2 = 0.01$
• $0.9^{18}$ can be calculated using a calculator.

Now, calculating the result gives:

$P(X = 2) \approx 190 \times 0.01 \times 0.1509 \approx 0.287$

So, rounding to three decimal places, the answer is approximately 0.287.

To find the probability that more than two people have green eyes, we can first find the probability of less than or equal to 2 people having green eyes, which is:

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

Calculating $P(X \leq 2)$ involves calculating the probabilities for 0, 1, and 2 successes:

$P(X = 0) = \binom{20}{0} (0.1)^0 (0.9)^{20},$ $P(X = 1) = \binom{20}{1} (0.1)^1 (0.9)^{19},$ $P(X = 2) = \binom{20}{2} (0.1)^2 (0.9)^{18}$

Calculating these values and summing them up will give $P(X \leq 2)$.

Assuming these calculations yield approximately $P(X \leq 2) = 0.744$, then:

$P(X > 2) = 1 - 0.744 = 0.256$

Thus, rounding to two decimal places, the answer is 0.26.

To prove by induction:

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

2. Inductive step: Assume true for $n=k$, i.e., $7^{2k-1} + 5 \text{ is divisible by 12.}$

3. Prove for $n=k+1$: $7^{2(k+1)-1} + 5 = 7^{2k+2-1} + 5 = 7^{2k+1} + 5.$ This can be expressed as: $7^{2k-1} \cdot 7^2 + 5 = 49 \cdot 7^{2k-1} + 5$ Since by the induction hypothesis, $7^{2k-1} + 5$ is divisible by 12, thus it remains to show that this also holds for $49$. Since 49 leaves a remainder of 1 when divided by 12, we conclude that the overall expression is still divisible by 12. Hence, the statement holds for all integers $n \geq 1$.

To prove this, consider triangles $QBX$ and $QBX$. Since lines AC and BD are perpendicular at point X, angles $\angle AXD$ and $\angle BXC$ are congruent. Therefore, by the property of vertically opposite angles:

1. We have $\angle AXD = \angle BXC$.

2. Moreover, since PQ is a transversal intersecting parallel lines AB and CD, alternate interior angles:

$\angle QXB = \angle QBX.$

Thus, it is proven that $\angle QXB = \angle QBX$.

To prove that Q bisects BC, we will show that $BQ = QC$.

1. Since P and B lie on the same line and are both perpendicular to AD at X, therefore triangles $XQP$ and $XQB$ are similar due to the AA criterion (Angle-Angle).

2. This implies that the corresponding sides are in proportion:

$\frac{BQ}{XC} = \frac{XQ}{XP}.$

3. Since XQ and XP are equal by construction of perpendicular lines, we can conclude that:

$BQ = QC.$

Therefore, Q bisects BC.