If I have the 5 colored shapes shown that I can rotate, and I use each shape once, is it possible to place them so they fit perfectly in a \(5 \times 4\) rectangle?

**Hint**: Notice the checkerboard pattern of the rectangle, and consider how the shapes would be colored if they followed the same pattern.

Friends Stella and Robert each have a ball of the same size, except Stella's ball is made of steel, while Robert's is rubber. They are having a competition to see who can launch their ball farthest. If they each have a cannon capable of accelerating their ball to \(50\text{ m/s}\) at the same angle of projection, who will win?

**Details and Assumptions:**

- The balls are shot through the air, but no wind is blowing.
- The distance is measured from the cannon to the point at which the ball first hits the ground.

\[\begin{align} \underbrace{\dfrac23 + \dfrac23 + \cdots + \dfrac23}_{\color{green}{A}\color{black} \text{ copies of } \frac23} &= \underbrace{\dfrac45 + \dfrac45 + \cdots + \dfrac45}_{\color{blue}{B}\color{black} \text{ copies of } \frac45} \\\\\\ 20 \leq \color{green}{A}\color{black}+\color{blue}{B}\color{black} &\leq 24\\\\ \color{green}{A}\color{black}+\color{blue}{B}\color{black} &=\, ? \end{align}\]

I have \(\color{green}{A}\color{black}\) copies of \(\frac23\) on the left-hand side of the equality, and \(\color{blue}{B}\color{black}\) copies of \(\frac45\) on the right. In total, I've written down between 20 and 24 fractions. Exactly how many fractions are there?

The birthday paradox is a surprising result of probability. Suppose you randomly chose 23 people and put them in a room. Then there would be a good chance \(\big(\)greater than \(\frac{1}{2}\big)\) that two of those people share a birthday (even though there are 365 days in the year).

What about birth seasons (spring, summer, fall, winter)? Suppose you randomly chose 3 people and put them in a room. Then is it true that there would be a greater than \(\frac{1}{2}\) chance that two of them share a birth season?

**Note**: Birth seasons do not all have the *exact* same likelihood. However, their likelihoods are close enough that you can assume they are equal for this problem.

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