We wanted to find the probability that the row started by a pear ends with one. Since no conditions limit the number of fruits, there are infinite solutions. Our answer is thus one third.
Solution to Fascinating Fruits
Month: June 2019
Fascinating Fruits
You will need an understanding of algebra and combinatorics to solve today’s challenge.
Today’s challenge begins with 3 rows of fruits. They begin with an apple, and orange, and a pear. The fruits behind the front are unknown. Directly behind any apple, there is an equal chance of a plum, a banana, or an orange. There is no other fruit directly behind an apple, but one apple is at the end of a row. Behind any plum is the same fruit that was in front of it. Directly behind an orange, there is a one third probability there are grapes. There is an equal probality of pears and blueberries, and 1 orange leads to a banana. For any banana there is an equal chance of blueberries or a plum. 1 banana is followed by grapes, and another ends a row. Behind blueberries, there is a one third chance of a pear, otherwise grapes. Directly behind grapes, there is a 50% chance of more grapes, a 25% chance of a plum, and otherwise a pear. Finally, directly behind a pear there is the fruit of the colour in the rainbow following that of the fruit in front of the pair. The exception is the 1 pear that ends a row. The apples in this question are red, which is considered to follow purple, the colour of the plums. The fruit following the pair at the front is an apple. The challenge is to use this information to find the probability that the row started by a pear ends with one.
Solution to Puzzling Probabilities
We wanted to find the probability, x, that Bob wins a game of Stupid Spin at Scam Casino. The probability Bob gets a point is a, otherwise Scam Casino gets one. The first to b points wins. First, the maximum spins in a round is 2b – 1, where 1 player wins by just 1 point. If we always play this many spins, you will always win at b points. We can simplify the problem to whomever has more points after 2b – 1 spins. Bob wins if the Casino has 0 to b-1 points. This gives us the first part of the equation. The probability of any given score for the Casino is equal to the number of ways that score can happen times the probability of it happening. The number of ways is easily done with the choose function. The probability of that is a to the power of the number of points scored by Bob, part three, multiplied by 1-a to the power of the number of points scored by the casino, the final part. We combine these 4 parts to get our answer, the equation below.
\sum_{k=0}^{b-1} {2b-1 \choose k}a^{2b-k-1}(1-a)^{k}Puzzling Probabilities
This challenge, though it appears simple, is far more difficult than you might expect. Knowledge of algebra, combinatorics, and graph theory, as well as ingenuity and effort, are necessary to successfully complete this challenge. Be warned.
Bob goes to Scam Casino to lose some money. He decides to play the game “Stupid Spin.” There are many equal pieces of the spinner, and only a few give Bob a point. Otherwise, the Casino gets one. The probability that Bob gets a point is a, and the first to b points wins. The challenge is to find the probability that Bob will win a given game of Stupid Spin.
Solution to Crazy Capture
We wanted to find the minimum number of police officers to capture a target. Remember that all units and the target are points with constant speed, perfect logic, and instant reaction time. Consider the lines perpendicular to those through the target and officers. Clearly, the target cannot cross any of these lines, the officers would always block it off. Furthermore, the target cannot delay all the officers in this way at once. The inactive ones can close in. It follows that a closed shape can catch the target. The smallest 2D shape is a triangle. We realize that we can generalize this. In higher dimensions, we use perpendicular planes and such. But the smallest figure is always the simplex! The simplex has n+1 vertices. Therefore, our answer is that 3 constables are needed to catch the target in 2 dimensions.
Crazy Capture
All of your algebra, geometry, trigonometry, angle theorems, and pathmaking skills are required to have a chance of solving today’s challenge.
In a mathematical capture, all police officers and the target are points that travel at the same constant speed in infinite space in n dimensions. They have infinite endurance and perfect reaction time. The target is instantly captured and cannot ecape when a police officer touches it. In 1 dimension, when n = 1, it obviously takes 2 officers surrounding the target. When n = 2, we can do it with three easily. We do this by dividing the plane into three, each officer can easily trap the target in a 120° slice. The challenge is to see how many officers it takes to capture the target when n equals 3, in three dimensions.
Solution to Devising Deactivation
We wanted to find a way for the invader to disable the radar undetected. Remember that Radar has a radius of 100 metres and rotates once every 12 seconds, and Invader runs at 7 m/s with their pack. Clearly, Invader waits until Radar has just passed it to begin. It takes Invader over 14 seconds to run straight to Radar. Maybe Invader could run a circle around Radar when it is about to catch up? For Invader yo circle Radar, they must be within 42/π, or 13.369, metres. That’s because the circle Invader runs must have radius 84 metres to be completed in 12 seconds. 7 m/s * 12 s = 84 m. This still leaves 86.631 metres left for Invader to run in 12 seconds, but they can only run 84 of them. So that’s a no. But what if Invader runs diagonally towards this circle? Invader then has 15 seconds to run √(1002+13.3692) = 100.89 metres. Invader can do 105! Invader does intersect the circle earlier on this path, so it’s not perfect, but it’s simple and works.
Devising Deactivation
To solve this challenge, you will need to use geometry, trigonometry, and algebra, to do pathmaking.
You are working for the government as a mathematician and they tell you of a problem. There is a plan afoot to compromise an enemy radar station. It is critical that the invading crew member is undedected, as the enemy must not know that their radar is sending them misinformation. However, the radar has a range of 100 metres and completes a rotation every 12 seconds. The fastest crew member can only sprint at 7 metres per second with their toolkit. Because of the radar’s length and rotation speed, it is nearly impossible to disable it undetected. None of the simple strategies work. The challenge is to find a path that the invader can take without being detected.