Question 1: Area Calculation

You are given a square with width = 1 and height = 1. Two quarter circles are drawn:

• One centered at the bottom-left corner
• One centered at the top-right corner

The shaded area is their overlapping region.

Area of the shaded region

Each quarter circle has radius 1 (same as the side of the square). The area of one quarter circle is:

(1/4) * π * r² = (1/4) * π * 1² = π / 4

The overlapping region is the intersection of the two quarter circles. This region is symmetric and appears as a lens-shaped area between them.

The total area of both quarter circles is:

π / 4 + π / 4 = π / 2

The area of the square is:

1 * 1 = 1

So the shaded overlapping region (intersection only) is:

2 * (Area of sector – Area of triangle formed between the arc endpoints and the center)

= 2 * ((π/4) – (1/2))
= (π/2) – 1

Answer:

(π / 2) – 1

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Question 2: Probability of a Flush

We draw 5 random cards from a 52-card deck. We want the probability of getting a flush, which is 5 cards of the same suit, excluding straight flushes.

Total number of 5-card hands

C(52, 5)

Ways to choose a flush (same suit):

• There are 4 suits.
• For each suit, we choose 5 cards: C(13, 5)
• So total flushes (including straight flushes): 4 * C(13, 5)

Subtract straight flushes:

• There are 10 possible straight flushes per suit (from A-5 to 10-A).
• So total straight flushes: 4 * 10 = 40

Valid flushes (excluding straight flushes):

4 * C(13, 5) – 40

Final probability:

P = [4 * C(13, 5) – 40] / C(52, 5)

Answer:

P = [4 * C(13, 5) – 40] / C(52, 5)

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Question 3: Egg Drop Puzzle (2 Marbles, 100 Floors)

You are given 2 marbles and a 100-floor building. You want to determine the highest floor from which you can drop a marble without breaking it, using the minimum number of drops in the worst case.

Optimal strategy

Use a decreasing interval strategy with the first marble:

• Drop the first marble from floors: x, x+(x–1), x+(x–1)+(x–2), ..., until it breaks.
• Then use the second marble to search linearly upward from the last safe floor.

The total number of drops is minimized when:

x + (x – 1) + (x – 2) + ... + 1 ≥ 100
=> x(x + 1)/2 ≥ 100

Solving:

x² + x – 200 ≥ 0

The smallest integer solution is x = 14, because:

14 * 15 / 2 = 105 ≥ 100

So the worst-case minimum number of drops required is:

Answer:

14 drops

Strategy Summary

1. Drop from floor 14, then 27 (14+13), then 39 (+12), etc.
2. When the first marble breaks, use the second marble to do linear search from the previous floor.
3. This minimizes the worst-case number of drops to 14.

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