Q 5.119 Archery. An archer shoots an arr... [FREE SOLUTION]

Chapter 5: Q 5.119 (page 225)

Archery. An archer shoots an arrow into a square target 6 feet on a side whose center we call the origin. The outcome of this random experiment is the point in the target hit by the arrow. The archer scores 10 points if she hits the bull's eye-a disk of radius 1 foot centered at the origin; she scores 5 points if she hits the ring with inner radius 1 foot and outer radius 2 feet centered at the origin; and she scores 0 points otherwise. Assume that the archer will actually hit the target and is equally likely to hit any portion of the target. For one arrow shot, let S be the score.
(a) Obtain and interpret the probability distribution of the random variable S. (Hint: The area of a square is the square of its side length; the area of a disk is the square of its radius times.)
(b) Use the special addition rule and the probability distribution obtained in part (a) to determine and interpret the probability of each of the following events:
${S = 5); {S > 0}; {S 鈮� 7); (5 < S 鈮� 15); {S < 15); and {S < 0) .$

Short Answer

Expert verified

Part (a)

$P (S = 10) = \frac{蟺}{36} P (S = 5) = \frac{蟺}{12} P (S = 0) = 9 - \frac{蟺}{9}$

Part (b)

$P (S = 5) = \frac{蟺}{12} P (S > 0) = \frac{4 蟺}{36} P (S 鈮� 7) = 36 - \frac{蟺}{36} P (5 < S 鈮� 15) = \frac{蟺}{36} P (S < 15) = 1 P (S < 0) = 0$

Step by step solution

Part (a) Step 1. Given information

10 points: disk with a radius of 1 foot

5 points: a ring with an inner radius of 1 foot and an outer radius of 2 foot

0 points: otherwise

Part (a) Step 2. Solution

Total Area of a square:

$A_{t o t a l} = {(6)}^{2} = 36$

For 10 points:

$A 10 p o i n t s = 蟺 {(1)}^{2} = 蟺$

For 5 points:

$A_{5 p o i n t s} = 蟺 {(2)}^{2} - 蟺 {(1)}^{2} = 3 蟺$

For 0 points:

$A_{0 p o i n t s} = A_{t o t a l} - A_{10 p o i n t s} - A_{5 p o i n t s} A_{0 p o i n t s} = 36 - 蟺 - 3 蟺 = 36 - 4 蟺$

Now,

role="math" localid="1651601043845" $P (S = 10) = \frac{A_{10 p o i n t s}}{A_{t o t a l}} = \frac{蟺}{36} P (S = 5) = \frac{A_{5 p o i n t s}}{A_{t o t a l}} = \frac{3 蟺}{36} = \frac{蟺}{12} P (S = 0) = \frac{A_{0 p o i n t s}}{A_{t o t a l}} = 36 - \frac{4 蟺}{36} = 9 - \frac{蟺}{9}$

Part (b) Step 1. Solution

$S = 10, S = 5 and S = 0$ are mutually exclusive events.

$P {S = 5} = \frac{蟺}{12} P {S > 0} = P {S = 5} + P {S = 10) = \frac{4 蟺}{36} P {S 鈮� 7} = P {S = 0} + P {S = 5} = 36 - \frac{蟺}{36} P {5 鈮� S 鈮� 15} = P {S = 10} = \frac{蟺}{36} P {S < 15} = P {S = 0} + P {S = 5} + P {S = 10} = 1 P {S < 0} = 0$