Question

Experiment II A sample space contains seven simple events: \(E_{1}, E_{2}, \ldots, E_{7} .\) Suppose that \(E_{1}, E_{2}, \ldots, E_{6}\) all have the same probability, but \(E_{7}\) is twice as likely as the others. Find the probabilities of the events. $$ A=\left\\{E_{3}, E_{4}, E_{6}\right\\} $$

Short answer

Answer: The probability of event A is \(\frac{3}{8}\).

Step-by-step solution

Assign variables to the probabilities

Let the probability of each of \(E_1, E_2, ..., E_6\) be \(p\). Then, the probability of \(E_7\) is \(2p\) because it is twice as likely as the others.

Sum the probabilities

We know that the sum of the probabilities of all the events in a sample space is equal to 1. So, we have: $$ p + p + p + p + p + p + 2p = 1 $$

Solve for p

Now, simplify the equation and solve for \(p\): $$ 8p = 1 $$ $$ p = \frac{1}{8} $$

Find the probabilities of all events

Now that we have the value of p, we can find the probabilities of all the events: $$ P(E_1) = P(E_2) = P(E_3) = P(E_4) = P(E_5) = P(E_6) = \frac{1}{8} $$ $$ P(E_7) = 2p = 2 \cdot \frac{1}{8} = \frac{1}{4} $$

Find the probability of event A

Event A is the set \(\{E_3, E_4, E_6\}\). To find the probability of event A, we add the probabilities of \(E_3, E_4,\) and \(E_6\): $$ P(A) = P(E_3) + P(E_4) + P(E_6) = \frac{1}{8} + \frac{1}{8} + \frac{1}{8} = \frac{3}{8} $$ The probabilities of the events are as follows: $$ P(E_1) = P(E_2) = P(E_3) = P(E_4) = P(E_5) = P(E_6) = \frac{1}{8}, \quad P(E_7) = \frac{1}{4}, \quad P(A) = \frac{3}{8} $$

Key concepts

Understanding Sample Space

Imagine you're about to roll a dice. Each side that could land face-up represents a possible outcome, collectively forming a 'sample space' in probability. In our problem, the sample space consists of seven simple events, labeled as \(E_{1}\), \(E_{2}\), and so on until \(E_{7}\). The key is, all these events encapsulate every possible thing that could happen; nothing is left out.

Understanding the concept of sample space is crucial because it anchors the fundamental principle of probability—that all possible outcomes must be accounted for. When events like \(E_{7}\) have different probabilities (it's twice as likely), we must adjust our calculations to reflect this imbalance.

Representing Sample Space

Traditionally, a sample space is represented using a list, a table, or a diagram. For example, if you think about flipping two coins, your sample space would be \[\{HH, HT, TH, TT\}\] (where H stands for heads and T for tails), covering all possible combinations of the coins' outcomes.

Calculating Event Probability

The likelihood of an event occurring within our defined sample space is called 'event probability.' It's a fraction or a percentage representing the chances of that event happening. In the dice analogy, the probability of rolling a four is 1 out of 6 or \(\frac{1}{6}\).

In our textbook exercise, the probability of each event \(E_{1}\) through \(E_{6}\) is \(p\) and \(E_{7}\) has a probability of \(2p\) since it is twice as likely to occur as the others. Here's a real-life analogy: if you had a bag with six red balls (\(E_{1}\) through \(E_{6}\)) and one blue ball (\(E_{7}\)) that's magically twice as likely to be picked, you'd have to account for that increased likelihood in your probability calculations, just as we did for \(E_{7}\).

Factors Affecting Event Probability

Several factors can influence the probability of an event, including the number of possible outcomes, the nature of these outcomes (whether some are more likely than others), and the presence of conditions or rules that could skew the results (like in games where certain cards might be removed).

Solving Probability Problems

Solving probability problems often involves a series of logical steps guided by the principles of probability. First, identify your sample space and any unusual aspects, like if some events have different probabilities (as we saw with \(E_{7}\)). Next, calculate the probability of each event. Then, depending on the problem, you might sum these probabilities, just as we did to find the chance of event \(A\) (comprised of \(E_{3}, E_{4}, E_{6}\)).

To enhance this process:

• Always ensure your total probability across the sample space equals 1, reflecting the certainty that one of the outcomes will occur.
• Look for patterns or symmetries in the problem that might help simplify your calculations.
• Be comfortable with basic algebra to solve for unknown probabilities.
• Keep in mind the difference between independent events (like multiple coin tosses) and dependent events (like drawing cards without replacement).
• Consider drawing a probability tree or using a Venn diagram if the problem involves multiple steps or stages.

By following these steps and considering our advice, you'll be more capable of handling various probability problems and finding accurate solutions.