As a continuation we will consider two more puzzle which motivates one or more applications in our day to day life.
Solution:
Method 1 : Each chord of the circle is uniquely defined by its center. We can place the chord's center inside the equilateral triangle area, so that it always falls inside the interior of the triangle. Thus the chance of choosing a chord which is in the interior of the triangle is
\(\mathbb{P}(\text{chord} \in \Delta) = \frac{\text{Area of } \Delta}{\text{Area of Circle}} = \frac{3\sqrt{3}}{4 \pi}\)
Method 2 : Another method to express the probability is as follows; Fix one of the end points of the chord at the vertex of the equilateral triangle.. The chords that can fall in the interior will fall in the angle range \(\frac{\pi}{3} \leq \theta \leq \frac{2 \pi}{3}\) . Thus the probability of the event of interest
\(\mathbb{P}(\text{chord} \in \Delta) = \frac{\pi/3}{\pi} = \frac 13\)
Method 3 : The chord is defined by its distance from the center of the circle.Thus the chords with distance from the center less than \(r/2\) can be classified as the interior of the triangle. Thus the chance that the chord chosen at random will fall in the interior of the triangle is given as:
\(\mathbb{P}(\text{chord} \in \Delta) = \frac{r/2}{r} = \frac 12 \)
Notice the three distinct answers we obtained. Can you think of a good reason to justify this.
If we look into the situation more clearly it is evident that the notion of uniformly random chord is not well defined. In that case all 3 methods are right . Unless a fixed notion(reference) has been foretold, there is no clear solution. Such class of problems (challenges!) are called ill-posed .
Puzzle #1: A school math teacher wants to teach geometry to her class 8 students. She collects 2 hemisphere shells and a cuboid, such that the cuboid can be inscribed inside the shell. She also bought paints -red and blue. One hemisphere was painted red, while only 80% of the second shell could be finished with the red paint.. So she decided to paint the remaining by blue paint. The teacher put the cuboid inside the sphere, and sees that all vertices fall on the red colour (assume that the shell is semi-transparent). Will her observation be true irrespective of the way she had painted.
The video solution of the puzzle above : -
Solution: If we consider the two hemispherical shells as a sphere, we understand that 10% of the sphere is painted in blue. Let us now denote, the vertices of the cube as \(v1\), \(v2\), \(v3\), \(v4\), \(v5\), \(v6\), \(v7\) and \(v8\). The chance that a given vertex is placed in the blue part of the sphere, is given by:
\[\mathbb{P}(\cup_i v_i \in Blue) = 0.1 \times 8 = 0.8\]
Thus, the chance that all eight vertices fall on the red part of the sphere \(= 0.2\). This means that there is at least one way of arranging the vertices such that all will fall on the red paint. The assertion made by the teacher is right.
(Courtesy : NIT Lecture, IITB)
Another one from geometry
Puzzle #2: A statistician is interested in this particular geometrical issue to find a spectrum allotment. An equilateral triangle is inscribed in a circle. The altitude of the equilateral triangle is 3/2 r. Find the chance (probability) that a chord chosen at random will intersect with the interior of the equilateral triangle.
Try to work this out using more than one method. Is the answer you get the same in all cases?
Solution:
Method 1 : Each chord of the circle is uniquely defined by its center. We can place the chord's center inside the equilateral triangle area, so that it always falls inside the interior of the triangle. Thus the chance of choosing a chord which is in the interior of the triangle is
\(\mathbb{P}(\text{chord} \in \Delta) = \frac{\text{Area of } \Delta}{\text{Area of Circle}} = \frac{3\sqrt{3}}{4 \pi}\)
Method 2 : Another method to express the probability is as follows; Fix one of the end points of the chord at the vertex of the equilateral triangle.. The chords that can fall in the interior will fall in the angle range \(\frac{\pi}{3} \leq \theta \leq \frac{2 \pi}{3}\) . Thus the probability of the event of interest
\(\mathbb{P}(\text{chord} \in \Delta) = \frac{\pi/3}{\pi} = \frac 13\)
Method 3 : The chord is defined by its distance from the center of the circle.Thus the chords with distance from the center less than \(r/2\) can be classified as the interior of the triangle. Thus the chance that the chord chosen at random will fall in the interior of the triangle is given as:
\(\mathbb{P}(\text{chord} \in \Delta) = \frac{r/2}{r} = \frac 12 \)
Notice the three distinct answers we obtained. Can you think of a good reason to justify this.
If we look into the situation more clearly it is evident that the notion of uniformly random chord is not well defined. In that case all 3 methods are right . Unless a fixed notion(reference) has been foretold, there is no clear solution. Such class of problems (challenges!) are called ill-posed .
(Courtesy : Introduction to Probability Theory, IITB )
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