6. Worked Example

Problem: A single fair die is rolled. Let event A = “rolling an even number” = {2, 4, 6}. Let event B = “rolling a number ≥ 4” = {4, 5, 6}.

In this example, we are examining the outcomes of rolling a single fair die, which has six faces numbered from 1 to 6. Two events are defined for this problem: Event A is rolling an even number, which can be either 2, 4, or 6. Event B is rolling a number that is greater than or equal to 4, which includes 4, 5, and 6. These definitions help us to focus on specific results we are interested in when the die is rolled.

Imagine you’re playing a simple game where you roll a die to determine your next move in a board game. If you need to roll an even number to advance your piece, you will be hoping for a 2, 4, or 6. Alternatively, if you want to score points by rolling a number that is 4 or higher, you will be looking for 4, 5, or 6. This situation makes it clear what each event means in a fun context.

• 𝑃(𝐴) = 3/6 = 0.5

The probability of event A, which is rolling an even number, is calculated by determining the number of favorable outcomes divided by the total number of possible outcomes when rolling a fair die. There are three favorable outcomes (2, 4, 6), and there are six possible outcomes in total (1, 2, 3, 4, 5, 6). Therefore, the probability P(A) is 3 divided by 6, which simplifies to 0.5 or 50%.

Think of it like having a bag with 6 different colored marbles, where 3 are blue (even numbers). If you were to reach into the bag without looking, there's a 50% chance you'd pull out a blue marble because 3 of the 6 marbles are blue. This directly relates to our previous scenario of rolling even numbers.

• 𝑃(𝐵) = 3/6 = 0.5

Similar to our calculation for event A, the probability for event B, which is rolling a number greater than or equal to 4, is also calculated by identifying the number of favorable outcomes (4, 5, 6) over the total number of possible outcomes. There are again three favorable outcomes out of six; hence, P(B) is also 3 divided by 6, resulting in a probability of 0.5 or 50%.

Imagine you're in a contest where prizes are awarded for rolling a die and you want to score points for rolling a number 4 or higher. With three winning numbers (4, 5, 6) out of six, like flipping a coin that has a heads side, you have an equal chance of winning or losing in this scenario as well.

• 𝐴∩𝐵 = {4,6}, so 𝑃(𝐴∩𝐵) = 2/6 = 1/3

The intersection of events A and B (A ∩ B) means that we are looking for outcomes that satisfy both conditions: rolling an even number and rolling a number greater than or equal to 4. The outcomes that fit both criteria are 4 and 6. Thus, there are 2 favorable outcomes. Therefore, the probability of both events occurring together (the intersection) is 2 divided by the total 6 outcomes, resulting in P(A ∩ B) = 2/6 = 1/3.

Let’s say you are rolling a die once and trying to fulfill two game requirements at the same time: you want an even number, but it also must be a high enough (4 or more) to gain a bonus. The only numbers that meet both criteria are 4 and 6, leaving you with two success chances out of a total of six.

• 𝐴∪𝐵 = {2,4,5,6}, so 𝑃(𝐴∪𝐵) = 4/6 = 2/3

The union of events A and B (A ∪ B) includes outcomes that are in either event A or event B or in both. In this case, the outcomes from event A are {2, 4, 6} and from event B are {4, 5, 6}. Combining these unique outcomes gives us {2, 4, 5, 6}, totaling four successful outcomes. Therefore, P(A ∪ B) = 4 divided by the total 6 outcomes, resulting in 4/6 = 2/3.

Picture a buffet where you can choose from 6 different dishes (the die), and you can satisfy two different cravings (even number or a number greater than 4). If you consider both cravings together, you can choose 2, 4, 5, or 6 to fill your plate, giving you more options to enjoy your meal, which translates into a higher success rate.

• Check addition: 0.5+0.5−1/3 = 1−1/3 = 2/3; matches.

To verify our calculations, we apply the addition rule of probability, which states that the probability of either event A or event B occurring is equal to the sum of their individual probabilities minus the probability of both events occurring together. In this case, we take P(A) + P(B) - P(A ∩ B) = 0.5 + 0.5 - 1/3. To simplify, convert to a common denominator of 3: 1 - 1/3 = 2/3, which corresponds with our calculated probability for the union of A and B.

Imagine checking your scores in a game where you earn points for either achieving a certain score alone or meeting two conditions simultaneously. By using the addition method, you ensure you're not counting points for the same scenario twice when both tasks are achieved, helping you accurately determine your final score.

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