UW-Stevens Point Math Club
Problem of the Week Has Been Discontinued
The problem of the week is no longer an active project of the Math Club. Below is a copy of the original problems and solutions. We hope that you will find this a valuable resource.
Problem #3 and Solution
Compare Them
De Mere, an acquaintance of Pascal’s gave him the following problem for which Pascal provided a solution.
Which is greater, the probability of getting at least one six with four rolls of a single die or the probability of getting at least one double-six in 24 rolls of a pair of dice?
Show your work on this problem.
Solution
The probability of getting at least one six is one minus the probability of getting none.
The probability of getting at least one twelve is one minus the probability of getting none.
So the probability of getting at least one six in four rolls is greater.
Congratulations to Jeff Schumacher, the first to submit a correct solution!
Problem #2 and Solution
Color the Corners
Starting with 4 colors, say red, white, blue, and green, how many ways can the corners (vertices) of a square be colored? Note a vertex is dimensionless so that it can’t actually be “colored.” It is more proper to say we are assigning colors to the vertices. Assume that reflections and rotations are allowed (so some back views are equivalent to some front views, for example). Show your work on this problem.
Solution
We may divide the problem up using cases:
- Case 1: Using any particular one of the colors, say blue, the vertices can be colored in only one way. Thus using all, they can be colored in 4 ways.
- Case 2: Using two particular colors, say red and blue together where both must be used, the vertices can be colored in only four ways (rbbb, brrr, bbrr, brbr). Using all possible color combinations, there are thus 6(4) = 24 ways.
- Case 3: Using three particular colors, say red, white, and blue, there are only six ways the vertices can be colored where all three must be used (rrwb, rwrb, wwrb, wrwb, bbrw, brbw). Using all possible color combinations, there are thus 4(6) = 24 ways.
- Case 4: Using all four colors, where all must be used, there are 3 ways the vertices can be colored (rwbg, rbwg, rbgw).
Adding the values from these four cases, we get 55 ways total.
Problem #1 and Solution
Find the Probability
Ten poker chips have numbers written on them so that one chip is numbered “1,” two chips are numbered “2,” three chips are numbered “3,” and so on. The chips are placed in a bag and three chips are drawn at random and without replacement. What is the probability that the sum of the numbers on the chips drawn is divisible by 5?
Solution
The possible draws that allow division by 5 are given by:
- 1, 2, 2, in any order (with a sum of 5)
- 2, 4, 4 in any order (with a sum of 10)
- 3, 3, 4 in any order (with a sum of 10)
There are three possibilities of ordering each of the draws. The following is a calculation of the probability.
Congratulations Andrea, on having the first correct solution!
Problem #4 and Solution
Circle in Sector
A circle is inscribed in a circular sector with radius R and central angle θ (in radians) as shown. If c is the circumference of the circle and s is the arc length of the circular sector, determine the limit, as θ approaches 0 from the positive direction, of c divided by s.
Solution
The solution may be found here.
Congratulations Physics Club, on having the first correct solution!
Problem #3 and Solution
Circle in Triangle
Take an equilateral triangle with side length s. Inscribe a circle in this triangle. Find the area of the circle in terms of the side length s, and find the distance p from the top of the circle to the top of the triangle in terms of s. Is there an interesting fact about p?
Solution
The area is equal to (π/12)s^2. And p is equal to the radius of the circle, which is equal to s/(2√3).
Congratulations Joshua Hastey, on having the first correct solution!
Problem #2 and Solution
The Cow
A cow is tied to a silo with radius r by a rope just long enough to reach the opposite side of the silo. Find the area available for grazing by the cow.
Hint: The length of the rope is πr. Also draw a picture representing the maximum perimeter the cow can travel (think parametric with parameters r and θ).
Solution
The solution may be found here.
Congratulations Joshua Hastey, on having the first correct solution!