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What is the probability that A and B meet within 4:00–5:00?

The probability is 7/167/16. This is calculated using the geometric probability model: the area of the meeting region (the diagonal band) divided by the area of the sample space (the unit square). The area of the band is found by subtracting the areas of two corner triangles from the square's area.

Conditions

  • Arrival times are independent and uniform on [0,1][0,1].
  • The meeting condition is ∣x−y∣≤1/4|x-y| \le 1/4.
  • The sample space is the unit square with area 1.

Reasoning, step by step

  1. Calculate the area of the sample space: 1×1=11 \times 1 = 1.
  2. Identify the complement region: two right triangles at corners (0,1)(0,1) and (1,0)(1,0).
  3. Calculate the leg length of each triangle: 1−1/4=3/41 - 1/4 = 3/4.
  4. Calculate the area of one triangle: 12⋅(3/4)2=9/32\frac{1}{2} \cdot (3/4)^2 = 9/32.
  5. Calculate the total complement area: 2⋅9/32=9/162 \cdot 9/32 = 9/16.
  6. Subtract from total area: 1−9/16=7/161 - 9/16 = 7/16.
  7. Conclude P(A)=7/16P(A) = 7/16.

Example

The script states: 'Substituting into the video’s ratio gives P(A)=(1−(3/4)(3/4))/(1⋅1)=7/16P(A)=(1-(3/4)(3/4))/(1\cdot1)=7/16. This answer belongs to the independent uniform arrival model...'

Common misconceptions

  • Assuming the probability is 1/41/4 based on the time difference.
  • Forgetting to square the leg length in the triangle area formula.
  • Not subtracting the complement area from 1.

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