Chapter 2.2 Problems 2 and 12

Problem 2

Statement

Suppose you choose a real number \(X\) from the interval \([2, 10]\) with a density function of the form \[f(x) = Cx \text{ where } C \text{ is a constant}\]

  1. Find \(C\).
  2. Find \(P(E)\) where \(E = [a,b]\) is a subinterval of \([2,10]\).
  3. Find \(P(X>5)\) , \(P(X < 7)\), and \(P(X^2 - 12X + 35 > 0 )\).

Solution

Part a.

To find \(C\), we require the probability over the entire sample space to sum to one. \[\int_{2}^{10} f(x)dx = 1\] \[\int_{2}^{10} f(x) dx = \int_{2}^{10} Cx dx = \frac{C}{2}(10^2 - 2^2 ) = \frac{C}{2}(100-4)=1\]

This implies \[C = \frac{2}{96} = \frac{1}{48}\]

Part b.

Using the integral calculated in part a, we obtain

\[P(E) = \int_{a}^{b} f(x)dx = \frac{C}{2}( b^2 - a^2) = \frac{1}{96}(b^2 - a^2 )\]

Part c.

Using the formula in part b, we obtain

\[P(X>5) = P([5,10]) = \frac{1}{96}( 100 - 25) = \frac{75}{96} = \frac{25}{32}\]

\[P(X<7) = P([2,7]) = \frac{1}{96}( 49 - 4) = \frac{45}{96} = \frac{15}{32}\]

\[P(X^2 - 12X + 35 > 0 ) = P((X-7)(X-5) > 0 ) = P( X > 7 \text{ or } X < 5 ) = P(X>7) + P(X<5) = (1-P(X<7))+(1-P(X>5))\]

This implies that \[P(X^2-12X + 35 > 0) = (1 - \frac{15}{32}) + (1 - \frac{25}{32}) = \frac{24 }{32}=\frac{3}{4}\]

Problem 2

Statement

Take a stick of unit length and break it into 3 pieces, choosing the break points at random. The break points are assumed to be chosen simultaneously. What is the probability that the 3 pieces can be used to form a triangle? Hint: Use Exercise 8g.

Solution

To choose two break points, we require choosing values \(x,y\) independently from a unit interval \([0,1]\). This is equivalent to choosing the point \((x,y)\) from the unit square \([0,1]^2\).

Assume that \(x < y\), then we will find the solution set in that upper triangle of the unit square. The triangular inequalities require that 2 sides be longer than the third side. This gives 3 inequalities:

\[x + (y - x) > (1-y) \implies y > \frac{1}{2}\] \[x + (1-y) > (y-x) \implies y - x < \frac{1}{2}\]

\[ (y-x ) + (1 - y) > x \implies x < \frac{1}{2}\] These conditions are necessary and sufficient to have a triangle when $x < y $.

The region \(y > 1/2\) and \(x < 1/2\) is the upper left quadrant of the unit square. The constraint \(y - x < 1/2\) defines the lower triangle of that upper left quadrant. So it has area equal to \[\frac{1}{2} \cdot \frac{1}{4} = \frac{1}{8}\].

By symmetry, another solution set exists in the lower triangle as well also of area \[\frac{1}{8}\].

The combined probability of forming a triangle is \(\frac{1}{4}\) in the region defined by two subtriangles.

We plot the region using Monte Carlo simulation to generate \(n=10000\) dots and show that subset of points which define breaks that form a triangle.

n = 10000
x = runif(n)
y = runif(n)
is_triangular = ( ( x < 1/2) & ( y > 1/2 ) & (y - x < 1/2)) | (( x > 1/2 ) & (y < 1/2) & (x - y < 1/2))
plot(x[is_triangular], y[is_triangular])

(triangular_frequency = sum( is_triangular ) / length( is_triangular ) )
## [1] 0.2497

The empirical frequency of points that form a triangle is almost 0.25.