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A single fair die is tossed. Assign probabilities to the simple events and calculate the probabilities. \(B\) : Observe an even number

Short Answer

Expert verified
Answer: The probability of observing an even number when a single fair die is tossed is 1/2.

Step by step solution

01

Assign probabilities to the simple events

For a fair die, each face has an equal probability of showing up. So, we can assign probability \(\frac{1}{6}\) to each face, i.e., \(\{1, 2, 3, 4, 5, 6\}\).
02

Define event \(B\)

We are asked to calculate the probability of observing an even number, which we call event \(B\). The event \(B\) is the set of outcomes that are even numbers from the simple events: \(B = \{2, 4, 6\}\).
03

Calculate the probability of event \(B\)

To calculate the probability of event \(B\), we need to sum the probabilities of the simple events in the set \(B\). So, we have: \(\text{Probability of }B = P(2) + P(4) + P(6)\) As we have assigned equal probabilities to each face of the die in step 1, we have: \(\text{Probability of }B = \frac{1}{6} + \frac{1}{6} + \frac{1}{6} = \frac{3}{6}\)
04

Simplify the probability of event \(B\)

We can simplify the probability of event \(B\) by dividing the numerator and denominator by their greatest common divisor, which is 3. So, we have: \(\text{Probability of }B = \frac{3}{6} = \frac{1}{2}\) Therefore, the probability of observing an even number when a single fair die is tossed is \(\frac{1}{2}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Fair Die
When we talk about a fair die, we refer to a six-sided cube with numbers ranging from 1 to 6. Each side of the die is equally likely to land face up when rolled. This means that every side has an equal chance of appearing. Fairness ensures that no number is more likely to come up than another.

When rolling a fair die, the probability of any specific face showing up is \( rac{1}{6}\). This probability distribution is important for understanding outcomes and ensures the integrity of probability calculations.

In simpler terms, imagine if you were playing a game and you wanted each number to have an equal chance to win. A fair die guarantees just that by providing an identical opportunity for 1, 2, 3, 4, 5, or 6 to be rolled.
Simple Events
In probability, a simple event is an outcome that cannot be broken down further. For a die, simple events are the results of a single roll. Each roll of a six-sided die results in one of six possible outcomes: \(\{1, 2, 3, 4, 5, 6\}\).

These outcomes are mutually exclusive, meaning that when you roll the die, only one of these numbers can appear at a time.

Understanding simple events is crucial because they are the building blocks of more complex probability events. When assigning probabilities, we carefully consider each simple event and ensure they collectively cover all possible scenarios in the situation described.
Even Number
An even number is any integer divisible by 2 without a remainder. In the context of our die, the numbers 2, 4, and 6 are even. These are part of what we call event B, which refers to rolling an even number.

When dealing with a fair die, even numbers represent half the possible outcomes. As a result, calculating probabilities involving even numbers requires only considering these specific outcomes.

Even numbers have practical significance in probability exercises as they frequently form the basis of event-specific calculations. Events defined by even numbers help illustrate simple yet effective ways of calculating broader probability outcomes from basic characteristics.
Event Probability
The probability of an event is a measure of how likely that event is to occur. It is calculated by adding the probabilities of all the simple events that comprise the event.

In our case with event B, we calculate the probability of rolling an even number by summing the probabilities of rolling a 2, 4, or 6. Given each face of the die has a probability of \( rac{1}{6}\), the probability of event B becomes: \[P(B) = P(2) + P(4) + P(6) = rac{1}{6} + rac{1}{6} + rac{1}{6} = rac{3}{6} = rac{1}{2}\]
Event probability requires not only knowing the number of favorable outcomes but understanding how they relate to the total number of possible outcomes. This foundational principle of probability ensures that calculations are accurate and meaningful.

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Most popular questions from this chapter

Two cold tablets are unintentionally put in a box containing two aspirin tablets, that appear to be identical. One tablet is selected at random from the box and swallowed by the first patient. The second patient selects another tablet at random and swallows it. a. List the simple events in the sample space \(S\). b. Find the probability of event \(A\), that the first patient swallowed a cold tablet. c. Find the probability of event \(B\), that exactly one of the two patients swallowed a cold tablet. d. Find the probability of event \(C,\) that neither patient swallowed a cold tablet.

A group of research proposals was evaluated by a panel of experts to decide whether or not they were worthy of funding. When these same proposals were submitted to a second independent panel of experts, the decision to fund was reversed in \(30 \%\) of the cases. If the probability that a proposal is judged worthy of funding by the first panel is \(.2,\) what are the probabilities that: a. A worthy proposal is approved by both panels. b. A worthy proposal is disapproved by both panels. c. A worthy proposal is approved by one panel.

Two city council members are to be selected from a total of five to form a subcommittee to study the city's traffic problems. a. How many different subcommittees are possible? b. If all possible council members have an equal chance of being selected, what is the probability that members Smith and Jones are both selected?

A woman brought a complaint of gender discrimination to an eight-member HR committee. The committee, composed of five females and three males, voted \(5-3\) in favor of the woman, the five females voting for the woman and the three males against. Has the board been affected by gender bias? That is, if the vote in favor of the woman was \(5-3\) and the board members were not biased by gender, what is the probability that the vote would split along gender lines (five females for, three males against)?

In how many ways can you select five people from a group of eight if the order of selection is important?

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