Seeta has 5 mint and 4 chocolate favoured sweets wrapped alike. He chooses two sweets at random and eats them one after the other. Find the probability that Seeta ate sweets of different flavours.

Number of mint sweets = 5
Number of chocolate sweets = 4
Total number of sweets = $5 + 4 = 9$

There are two possible scenarios:

• Scenario 1: First sweet is mint, second sweet is chocolate (MC). Probability of picking a mint sweet first = $\frac{5}{9}$ After picking one mint sweet, there are 8 sweets left (4 mint, 4 chocolate). Probability of picking a chocolate sweet second = $\frac{4}{8}$ Probability of MC = $\frac{5}{9} \times \frac{4}{8} = \frac{20}{72}$

• Scenario 2: First sweet is chocolate, second sweet is mint (CM). Probability of picking a chocolate sweet first = $\frac{4}{9}$ After picking one chocolate sweet, there are 8 sweets left (5 mint, 3 chocolate). Probability of picking a mint sweet second = $\frac{5}{8}$ Probability of CM = $\frac{4}{9} \times \frac{5}{8} = \frac{20}{72}$

The probability that Seeta ate sweets of different flavours is $\boxed{\frac{5}{9}}$.

Triangle A has vertices at $(2, 4)$, $(5, 2)$, and $(8, 4)$.
Triangle B has vertices at $(2, -2)$, $(5, 0)$, and $(8, -2)$.

• $(2, 4)$ maps to $(2, -2)$
• $(5, 2)$ maps to $(5, 0)$
• $(8, 4)$ maps to $(8, -2)$

Notice that the x-coordinates remain unchanged, while the y-coordinates are transformed. This suggests a reflection across a horizontal line.

For a reflection across the line $y=k$, a point $(x, y)$ maps to $(x, 2k-y)$.
Let's use the first pair of points: $(2, 4) \to (2, -2)$.
So, $2k - 4 = -2$.
$2k = 2$.
$k = 1$.
The line of reflection is $y=1$.

Let's verify with another point: $(5, 2) \to (5, 2(1)-2) = (5, 0)$. This matches.

The single transformation that maps A onto B is a reflection in the line $y=1$.

A stretch with the y-axis invariant means that points on the y-axis (where $x=0$) do not change their position. The stretch factor of $\frac{1}{2}$ means that the x-coordinate of every point is multiplied by $\frac{1}{2}$, while the y-coordinate remains the same.
The transformation rule is $(x, y) \to (\frac{1}{2}x, y)$.

The vertices of triangle B are $(2, -2)$, $(5, 0)$, and $(8, -2)$.

• For $(2, -2)$: $(\frac{1}{2} \times 2, -2) = (1, -2)$
• For $(5, 0)$: $(\frac{1}{2} \times 5, 0) = (2.5, 0)$
• For $(8, -2)$: $(\frac{1}{2} \times 8, -2) = (4, -2)$

To draw the image of B, plot the new vertices:

• $(1, -2)$
• $(2.5, 0)$
• $(4, -2)$ Connect these three points with straight lines to form the stretched triangle.