Question

A large wooden floor is constructed with wood-planks which are 2 inches wide. A uniform circular disk of diameter 2.25 inches is dropped at random on the floor. What is the probability that the disk touches three of the wood-planks? Give your answer as a decimal rounded to four places (i.e. X.XXXX).

Answer 1

The probability that the disk touches three of the wood-planks is 1.5105

To determine the probability that the circular disk touches three wood-planks when dropped randomly on the floor, let's analyze the scenario.

The disk will touch three planks if its diameter aligns such that it overlaps with the edges of three planks but doesn't extend beyond that. The total area covered by three planks will be 3× 2 = 6 inches

The area of the circular disk is π × (2.25/2)²

= π × 1.125²

= π × 1.265

= 3.9761 inches

The probability that the disk touches three of the wood-planks is given as;

= Area of three- plank region/Total area

= 6/ π × 1.125²

= 6/3. 9761

= 1.5105

Answer 2

Explanation:

The side length of each wooden plank is 2 inches.

Area of each wooden plank is

Area = 2² = 4 inches²

Area of 3 wooden planks is 3 × 4 = 12 inches²

The formula for determining the area of the circular disc is expressed as

Area = πr²

π = 3.14

Radius, r = diameter/2 = 2.25/2 = 1.125 inches

Area of circular disc = 3.14 × 1.125² = 3.974

Probability is expressed as

Number of favorable outcomes/number of total outcomes

Therefore, the probability that the disk touches three of the wood-planks is

3.974/12 = 0.3312