© All Rights reserved @ LearnWithDash
Step 1: Interpret the Event
We are selecting two real numbers from the interval [0,60] , say x and y .
The event A is that the absolute difference between these two numbers is at most a ,
i.e., |x - y| \le a .
Step 2: Identify the Sample Space
Geometrically, each pair (x, y) corresponds to a point in the square [0,60]\times[0,60] .
The total area of this square is 60 \times 60 = 3600 .
Since x and y are chosen uniformly, probability corresponds to
(area of the favorable region) ÷ (total area).
Step 3: Calculate the Area of the Favorable Region
The condition |x-y| \le a means that the point (x,y) lies within a band around
the line y = x , having width a above and a below this diagonal.
The well-known area of this band in the square of side L = 60 is given by:
\text{Area} = L^2 - (L - a)^2 = 60^2 - (60 - a)^2 = 3600 - (3600 -120a + a^2)
= 120a - a^2.
Step 4: Use the Given Probability to Form an Equation
We are told
P(A) = \frac{11}{36}.
Therefore,
\frac{\text{Area of favorable region}}{\text{Total area}} = \frac{11}{36}.
Substituting the expressions for the areas:
\frac{120a - a^2}{3600} = \frac{11}{36}.
Step 5: Solve for a
Clear the denominators by multiplying both sides of the equation by 3600:
120a - a^2 = 3600 \times \frac{11}{36} = 100 \times 11 = 1100.
Rearrange into standard quadratic form:
-\,a^2 + 120a - 1100 = 0 \quad \Longrightarrow \quad a^2 - 120a + 1100 = 0.
Solve this quadratic using the formula a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} ,
where a=1 , b=-120 , and c=1100 :
a
= \frac{120 \pm \sqrt{(-120)^2 - 4 \cdot 1 \cdot 1100}}{2}
= \frac{120 \pm \sqrt{14400 - 4400}}{2}
= \frac{120 \pm \sqrt{10000}}{2}
= \frac{120 \pm 100}{2}.
Hence the roots are:
a = \frac{120 + 100}{2} = 110
\quad \text{and} \quad
a = \frac{120 - 100}{2} = 10.
Since a is a maximum absolute difference within the interval [0,60] ,
a cannot exceed 60. Therefore,
a = 10
is the valid solution.
Final Answer
The value of a is \boxed{10} .
