We are given a system of inequalities. We want to find the probability that if a point (x, y) is chosen at random in solution set of the system, it is a solution to the inequality (x-1)^2 +(y-1)^2≥ 16. First, let's take a look at the given system.
1 ≤ x ≤ 6 y ≤ x y≥ 1
Since the first inequality is compound, we can rewrite it as two separate inequalities.
x ≤ 6 x ≥ 1 y ≤ x y≥ 1
Let's graph all the inequalities in one coordinate system.
Now, let's also graph the inequality (x-1)^2 +(y-1)^2≥ 16. The equation of the boundary line of this inequality is an equation of a circle with center (1,1) and radius 4 in standard form.
We can see that the intersection of the circle and the triangle is a sector of the circle. To find the probability that the point satisfies the given inequality, we will use geometric probability. The probability that a point chosen at random in the triangle satisfies the given inequality is the ratio of the area A of the region of the triangle outside the circle to the area A_T of the triangle .
P((x-1)^2+(y-1)^2 ≥ 16 ) = A/A_T
First, let's find the area of the triangle. We know the formula for the area A_T of a right triangle with legs of length a and b.
A_T = 1/2 a b
In our case, both legs have a length of 5. By substituting 5 for both a and b in this formula, we can calculate the area A_T of our triangle.
To find the area A of the region of the triangle that is outside of the circle, we will subtract the area A_S of the sector of our circle from the area A_T of the triangle. Recall the formula for the area of a sector of a circle with measure θ and radius r.
A_S = θ/360 π r^2
In our case, the measure of the sector is 45^(∘) and the radius is 4. By substituting these values for θ and r, respectively, in the formula, we can calculate the area A_S of the sector.
The probability that a point (x,y) chosen randomly in the solution set of the system of inequalities is a solution to the inequality (x-1)^2+(y-1)^2 ≥ 16 is approximately 0.496, or approximately 49.6 %.
