Manish has to travel from A to D changing buses at stops B and C enrou...
Understanding the Problem
Manish has a maximum waiting time of 8 minutes at each of the two stops (B and C) and he can afford a total waiting time of 13 minutes. This means:
- Waiting at Stop B: X (0 to 8 minutes)
- Waiting at Stop C: Y (0 to 8 minutes)
The total waiting time is given by the equation:
X + Y > 13
We need to find the probability that Manish arrives late at D.
Setting Up the Scenario
- Both X and Y are uniformly distributed between 0 and 8.
- The total possible waiting time combinations (X, Y) form a square region in the XY-plane with vertices (0,0), (8,0), (0,8), and (8,8). The area of this square is 64 (8*8).
Finding the Area of Interest
1. Late Arrival Condition:
- The condition X + Y > 13 represents a line on the XY-plane.
2. Identifying the Intersection:
- The line intersects the boundaries of the square at points (5,8) and (8,5).
3. Calculating the Area Above the Line:
- The triangle formed by these points and the origin (0,0) has vertices (5,8), (8,5), and (8,8).
Calculating Areas
- Area of the triangle:
- Base = 8 - 5 = 3
- Height = 8 - 5 = 3
- Area = 0.5 * base * height = 0.5 * 3 * 3 = 4.5
- Total area above the line = Area of the square - Area of the triangle = 64 - 4.5 = 59.5
Probability Calculation
- Probability of arriving late (X + Y > 13):
P = Area above the line / Total area of the square = (64 - 4.5) / 64 = 59.5 / 64 = 8/13
Conclusion
Thus, the probability that Manish will arrive late at D is 8/13, making option 'A' the correct answer.