Problem 16 - Chemistry The Molecular Science

Question

Suppose you make a tetrahedron and put numbers \(1,2,3,\) and 4 on each of the four sides. You toss the tetrahedron in the air and observe it after it comes to rest. (a) What is the probability that the tetrahedron will come to rest with the numbers \(2,3,\) and 4 visible? (b) What is the probability that the tetrahedron will come to rest with the numbers \(1,2,\) and 3 visible? (c) If you toss the tetrahedron 100 times, what is the most likely number of times you will see a 1 after it comes to rest?

Step-by-Step Solution

Answer

(a) \( \frac{1}{4} \); (b) \( \frac{1}{4} \); (c) 75 times.

Step 1: Determine the Total Number of Outcomes

The tetrahedron has four faces, each with a number: 1, 2, 3, and 4. When you toss the tetrahedron, one number is hidden (facing down), and three numbers are visible (on the sides). Therefore, there are 4 possible outcomes for how the tetrahedron might land, as each face could be facing down.

Step 2: Calculate the Probability for Part (a)

To find the probability of the numbers 2, 3, and 4 being visible, identify the number that must be facing down: it must be 1. Hence, there is only 1 favorable outcome (when face 1 is down), out of the 4 possible outcomes. Thus, the probability is \( \frac{1}{4} \).

Step 3: Calculate the Probability for Part (b)

To find the probability of the numbers 1, 2, and 3 being visible, identify the number that must be facing down: it must be 4. Hence, there is only 1 favorable outcome (when face 4 is down), out of the 4 possible outcomes. Thus, the probability is \( \frac{1}{4} \).

Step 4: Calculate Expected Number of Times Seeing '1' in 100 Throws

Every toss of the tetrahedron is independent from the others. The probability of landing such that '1' is visible is \( \frac{3}{4} \) (since '1' is visible in 3 out of 4 cases). In 100 throws, the expected number of times you see '1' is given by \( 100 \times \frac{3}{4} = 75 \).

Step 5: Conclusion

For part (a), the probability is \( \frac{1}{4} \). For part (b), the probability is also \( \frac{1}{4} \). For part (c), you will see the number 1 approximately 75 times out of 100 tosses.

Key Concepts

Probability calculation

Probability calculation is a fundamental concept in statistics used to determine how likely an event is to occur. It helps us assign a numerical value to uncertainty, ranging from 0 (impossible event) to 1 (certain event). In our tetrahedron exercise, we have four possible outcomes when the tetrahedron is tossed, meaning each face can potentially face down.

To calculate probability, we use the formula:

• Probability = \( \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} \)

Since there are four faces available, each possibility has an equal chance, hence a probability of \( \frac{1}{4} \). This simple calculation aids in determining how likely a particular configuration of faces (such as 2, 3, and 4 showing) is when tossed.

Random events

Random events are occurrences that cannot be predicted with certainty. They form the backbone of probability theory. Each time we toss the tetrahedron, it rolls and rests at a random position, which means the event is subject to chance.

To understand randomness, consider that:

• Outcomes are independent of one another.
• Each outcome is equally likely in a fair scenario.

The randomness ensures that the face landing down is unpredictable, contributing to all possible outcomes being equally possible until after the event has occurred. Observing these random phenomena is fundamental to practical applications of probability, whether it's predicting dice rolls or understanding the behavior of objects like our tetrahedron.

Expected outcomes

Expected outcomes help us predict the long-term behavior of random processes. In the context of tossing a tetrahedron, it involves calculating how often a particular face will show based on it being independent and identically distributed.

To compute the expected number of times a face like '1' will show up over 100 tosses, the formula used is:

• Expected Count = Total Throws \( \times \) Probability of a single face being up

In our case, the probability the number '1' is visible is \( \frac{3}{4} \) because it is visible in three out of four positions. Therefore, in 100 tosses, we'll likely see '1' approximately 75 times. This concept of expectation allows us to make informed predictions based on probabilities.