A ball selected at random from a bag containing 3 red balls and 4 black balls. The first ball is not replaced and a second ball is selected. Find the probability of selecting two different balls.

Total balls = 3 red + 4 black = 7 balls.

Scenario 1: First ball is Red (R), second ball is Black (B).
Scenario 2: First ball is Black (B), second ball is Red (R).

• Probability of drawing a Red ball first: $P(R_1) = \frac{\text{Number of red balls}}{\text{Total number of balls}} = \frac{3}{7}$.
• After drawing one red ball, there are 2 red balls and 4 black balls left, making a total of 6 balls.
• Probability of drawing a Black ball second (given the first was red): $P(B_2 | R_1) = \frac{\text{Number of black balls}}{\text{Remaining total balls}} = \frac{4}{6}$.
• Probability of Scenario 1: $P(R_1 \text{ and } B_2) = P(R_1) \times P(B_2 | R_1) = \frac{3}{7} \times \frac{4}{6} = \frac{12}{42}$.

• Probability of drawing a Black ball first: $P(B_1) = \frac{\text{Number of black balls}}{\text{Total number of balls}} = \frac{4}{7}$.
• After drawing one black ball, there are 3 red balls and 3 black balls left, making a total of 6 balls.
• Probability of drawing a Red ball second (given the first was black): $P(R_2 | B_1) = \frac{\text{Number of red balls}}{\text{Remaining total balls}} = \frac{3}{6}$.
• Probability of Scenario 2: $P(B_1 \text{ and } R_2) = P(B_1) \times P(R_2 | B_1) = \frac{4}{7} \times \frac{3}{6} = \frac{12}{42}$.

The probability of selecting two different balls is $\boxed{\frac{4}{7}}$.

3. a) Given that $\log_x 2\frac{1}{4} = 2$, calculate the value of $x$.

The definition of a logarithm states that if $\log_b a = c$, then $b^c = a$.
Applying this to the given equation:

Take the square root of both sides:

The base of a logarithm ($x$) must be positive and not equal to 1.
Therefore, $x = \frac{3}{2}$ is the valid solution.

The value of $x$ is $\boxed{\frac{3}{2}}$.

3. b) Prove that the exterior angle of a cyclic quadrilateral is equal to the interior opposite angle of the cyclic quadrilateral.

Let's extend side BC to a point E, forming an exterior angle $\angle DCE$.
We need to prove that $\angle DCE = \angle BAD$.

Angles $\angle BCD$ and $\angle DCE$ form a linear pair on the straight line BCE.
Therefore, their sum is $180^\circ$.

In a cyclic quadrilateral, the sum of opposite interior angles is $180^\circ$.
Angles $\angle BCD$ and $\angle BAD$ are opposite interior angles.
Therefore, their sum is $180^\circ$.

From equation (1), we can express $\angle DCE$ as:

From equation (2), we can express $\angle BAD$ as:

Since both $\angle DCE$ and $\angle BAD$ are equal to the same expression ($180^\circ - \angle BCD$), they must be equal to each other.

Thus, the exterior angle of a cyclic quadrilateral is equal to its interior opposite angle.