Question 313.
In the game of antidarts a player shoots an arrow into a rectangular board
measuring 6 metres by eight metres.
If the arrow is within one metre of the centre it scores 1 point,
between 1 and 2 metres away it scores 2,
between 2 and 3 metres away it scores 3,
between 3 and 4 metres and yet still on the board it scores 4,
further than 4 metres but still on the board it scores 5.
The arrows always land on the board but otherwise they are purely random.
Show that the probability that he scores exactly 15 points in 3 arrows is given by
(1 -(2/3)arcsin(3/4) - (1/8)sqrt7)^3.
Answer 313.
To score exactly 15 points in 3 arrows, each arrow must score 5 points
=> each arrow must land in the area of the board further than 4 meters from the center,
but still on the board which is the area shaded blue in the figure as given in the following link.
The equation of the circle with radius 4 having center at the center of the board and axes parallel to the rectangular edges of the board is given by
x^2 + y^2 = 16
=> area shaded green
= ∫ x dy ... (y=0 to 3)
= ∫ √(16 - y^2) dy ... (y = 0 to 3)
= [ (y/2) √(16 - y^2) + 8arcsin(y/4)] ... (y = 0 to 3)
= [(3/2) √7 + 8arcsin(3/4)]
=> area shaded blue
= area of the board - 4 * area shaded green
= 48 - 4 * [(3/2) √7 + 8arcsin(3/4)]
= 48 - 6√7 - 32arcsin(3/4)
=> Probability of an arrow scoring 5, P(5)
= area shaded blue / total area of the rectangle
= [48 - 6√7 - 32arcsin(3/4)]/(48)
= 1 - (1/8)√7 - (2/3)arcsin(3/4)
=> Probability of all the 3 arrows landing in blue area and thus scoring exactly a total of 15 points
= [1 - (1/8)√7 - (2/3)arcsin(3/4)]^3.
[Note: The data of arrows scoring 1, 2, 3 and 4 is redundant.]
Link to YA!
In the game of antidarts a player shoots an arrow into a rectangular board
measuring 6 metres by eight metres.
If the arrow is within one metre of the centre it scores 1 point,
between 1 and 2 metres away it scores 2,
between 2 and 3 metres away it scores 3,
between 3 and 4 metres and yet still on the board it scores 4,
further than 4 metres but still on the board it scores 5.
The arrows always land on the board but otherwise they are purely random.
Show that the probability that he scores exactly 15 points in 3 arrows is given by
(1 -(2/3)arcsin(3/4) - (1/8)sqrt7)^3.
Answer 313.
To score exactly 15 points in 3 arrows, each arrow must score 5 points
=> each arrow must land in the area of the board further than 4 meters from the center,
but still on the board which is the area shaded blue in the figure as given in the following link.
The equation of the circle with radius 4 having center at the center of the board and axes parallel to the rectangular edges of the board is given by
x^2 + y^2 = 16
=> area shaded green
= ∫ x dy ... (y=0 to 3)
= ∫ √(16 - y^2) dy ... (y = 0 to 3)
= [ (y/2) √(16 - y^2) + 8arcsin(y/4)] ... (y = 0 to 3)
= [(3/2) √7 + 8arcsin(3/4)]
=> area shaded blue
= area of the board - 4 * area shaded green
= 48 - 4 * [(3/2) √7 + 8arcsin(3/4)]
= 48 - 6√7 - 32arcsin(3/4)
=> Probability of an arrow scoring 5, P(5)
= area shaded blue / total area of the rectangle
= [48 - 6√7 - 32arcsin(3/4)]/(48)
= 1 - (1/8)√7 - (2/3)arcsin(3/4)
=> Probability of all the 3 arrows landing in blue area and thus scoring exactly a total of 15 points
= [1 - (1/8)√7 - (2/3)arcsin(3/4)]^3.
[Note: The data of arrows scoring 1, 2, 3 and 4 is redundant.]
Link to YA!
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