Question

For the experiments, list the simple events in the sample space, assign probabilities to the simple events, and find the required probabilities. A roulette wheel contains 38 pocketsthe numbers 1 through \(36,0,\) and \(00 .\) The wheel is spun and the "winning" pocket is recorded. If observing any one pocket is just as likely as any other, what is the probability of observing either 0 or \(00 ?\)

Short answer

Answer: The probability of observing either 0 or 00 on a roulette wheel is \(\frac{1}{19}\).

Step-by-step solution

Listing the Simple Events in Sample Space

In this case, the simple events are every potential outcome of spinning the roulette wheel, i.e., the pocket numbers. Our sample space, S, can be written as: \(S = \{1, 2, 3, ... , 36, 0, 00\}\).

Assign Probabilities to Simple Events

Each pocket on the roulette wheel is equally likely to occur, so the probability of any single event is the same for all pockets. Since there are a total of 38 pockets, the probability for each pocket can be calculated as: \(P(\text{single pocket}) = \frac{1}{38}\)

Calculate the Probability of Observing 0 or 00

We are interested in finding the probability of observing either 0 or 00. To do this, we need to find the total probability of observing these specific events. In probability theory, the probability of the union of two mutually exclusive events (events that cannot occur simultaneously) can be found by simply summing their individual probabilities: \(P(0 \text{ or } 00) = P(0) + P(00)\) Since the probability of observing each pocket is the same: \(P(0 \text{ or } 00) = \frac{1}{38} + \frac{1}{38} = \frac{2}{38}\)

Simplify the Probability

Now we simplify this fraction: \(P(0 \text{ or } 00) = \frac{2}{38} = \frac{1}{19}\) So, the probability of observing either 0 or 00 on the roulette wheel is \(\frac{1}{19}\).

Key concepts

Sample Space

In probability, the concept of a sample space is fundamental. A sample space is essentially the set of all possible outcomes in a given experiment. For the roulette wheel mentioned in the exercise, the sample space includes every number that the ball could land on.
To represent it formally, we denote the sample space as \( S \). For the roulette wheel, it consists of numbers 1 through 36, and includes 0 and 00 as well. Thus, the sample space can be written as:

• S = \( \{1, 2, 3, \ldots, 36, 0, 00\} \)

Each element in the sample space is called a simple event. Each simple event is mutually exclusive, meaning that if one occurs, the others cannot occur at the same time. Understanding this helps in calculating probabilities accurately, as it frames our expectation based on all possible results.

Mutually Exclusive Events

Mutually exclusive events are events that cannot take place at the same time. They are central when determining the likelihood of different outcomes. If one event happens, it excludes the possibility of the other. The question about the roulette wheel illustrates this concept well.
When the ball lands on a pocket labeled '0', it cannot simultaneously land on '00'. Thus, these events are mutually exclusive.

• If \( A \) is the event that the ball lands on '0'
• and \( B \) is the event that the ball lands on '00',

then \( A \) and \( B \) are mutually exclusive. In probability theory, when adding the probabilities of mutually exclusive events, you can simply sum their individual probabilities. This is because the occurrence of one excludes the occurrence of the other, making it straightforward to calculate summed probabilities.

Equally Likely Outcomes

In probability, equally likely outcomes describe a scenario where each possible outcome has the same probability of occurring. Equally likely outcomes simplify calculations because it means each outcome is as probable as any other.
For the roulette wheel, since it contains 38 pockets, each outcome (landing on any specific pocket) has a probability of \( \frac{1}{38} \). This implies that:

• Each pocket, whether it’s '1', '0', or '00', is equally likely to occur.
• No number has a higher chance of appearing than any other.
  

In cases like these, assigning probabilities becomes straightforward, as we just need to divide 1 by the total number of outcomes, in this case, 38.