Two points are chosen randomly on a line 9 cm long. Determine the probability that the distance between them is less than 3 cmCorrect answer is '0.5-0.6'. Can you explain this answer?

Question

Answer

Answer

x: The distance of first point from the start of the line

y: distance of second point from the start of the line segment

x,y ϵ[0,9]

So sample space is Area of region bounded by

x ≥ 0, y ≥ 0, x ≤ 9, y ≤ 9

This is square of side 9

Area = 81 cm²

The region of our interest is

|x-y| < 3

0 ≤ x ≤ 9

0 ≤ y ≤ 9

Area of shaded region = 2 (area of triangle) + area of rectangle

= 45

Probability = 45/81 =0.55

Answer

Problem:
Two points are chosen randomly on a line 9 cm long. Determine the probability that the distance between them is less than 3 cm.

Solution:
To solve this problem, we need to consider the possible positions of the two points on the line and determine the probability that the distance between them is less than 3 cm.

Understanding the problem:
We have a line with a length of 9 cm. We need to randomly choose two points on this line. The distance between these two points should be less than 3 cm.

Approach:
To determine the probability, we need to find the favorable outcomes (the number of ways the distance between the two points can be less than 3 cm) and the total outcomes (the total number of ways the two points can be chosen on the line).

Favorable outcomes:
The distance between the two points can be less than 3 cm in two cases:
1. When both points are within the first 3 cm of the line.
2. When one point is within the first 3 cm of the line and the other point is within the last 3 cm of the line.

Case 1:
In this case, both points are within the first 3 cm of the line. The probability of choosing the first point within the first 3 cm is 3/9, and the probability of choosing the second point within the first 3 cm is also 3/9. Therefore, the probability of this case is (3/9) * (3/9) = 1/9.

Case 2:
In this case, one point is within the first 3 cm of the line and the other point is within the last 3 cm of the line. The probability of choosing the first point within the first 3 cm is 3/9, and the probability of choosing the second point within the last 3 cm is also 3/9. Therefore, the probability of this case is (3/9) * (3/9) = 1/9.

Total outcomes:
The total number of ways the two points can be chosen on the line is the number of ways to choose 2 points from a line of length 9 cm. This can be calculated using the combination formula:

total outcomes = C(2, 9) = 9! / (2! * (9-2)!) = 36

Calculating probability:
To calculate the probability, we need to divide the favorable outcomes by the total outcomes:

probability = favorable outcomes / total outcomes = (1/9 + 1/9) / 36 = 2/9 / 36 = 2/9 * 1/36 = 2/324

Simplifying the fraction, we get:

probability = 1/162 ≈ 0.0062

Therefore, the probability that the distance between the two points is less than 3 cm is approximately 0.0062 or 0.62%.

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