Problem 15
Question
\(\boldsymbol{S}\) and \(\boldsymbol{T}\) are mutually exclusive events. Find \(\boldsymbol{P}(\boldsymbol{S} \text { or } \boldsymbol{T})\) $$ P(S)=\frac{3}{5}, P(T)=\frac{1}{3} $$
Step-by-Step Solution
Verified Answer
Therefore, the probability of event S or T occurring is 14/15
1Step 1: Identify The Given Information
From the problem statement, it can be gathered that P(S) = 3/5 and P(T) = 1/3, and S and T are mutually exclusive events.
2Step 2: Apply The Addition Rule For Mutually Exclusive Events
According to the 'Addition Rule' for mutually exclusive events, the probability of either event S or T occurring is given by: \( P (S \text{ or } T) = P(S) + P(T) \)
3Step 3: Substitute The Given Values
Substitute the given values of P(S) and P(T) into the equation: \( P(S \text{ or } T) = 3/5 + 1/3 \)
4Step 4: Simplify The solution
To simplify the answer, convert the fractions to have a common denominator and add: \( P (S \text{ or } T) = 9/15 + 5/15 = 14/15 \)
Key Concepts
Addition RuleProbabilityCommon DenominatorSimplification of Fractions
Addition Rule
When working with probabilities, especially in situations involving multiple events, it's crucial to determine how these events interact. This is where the Addition Rule comes in handy. The rule is particularly useful for determining the probability of one or more events occurring. Two types of addition rules exist:
- Addition Rule for Mutually Exclusive Events: This is applied when two events cannot happen at the same time, meaning the occurrence of one event precludes the occurrence of the other. For these events, the probability that either event A or event B will occur is simply the sum of their individual probabilities. Specifically, \( P(A \text{ or } B) = P(A) + P(B) \).
- Addition Rule for Non-Mutually Exclusive Events: When events can happen simultaneously, the rule is slightly different. You need to subtract the probability of both events occurring together: \( P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B) \).
Probability
The concept of Probability is a measure of the likelihood that a given event will occur. It ranges from 0 to 1, where 0 indicates impossibility and 1 indicates certainty. In practical terms, probability is calculated as follows:
- Probability Formula: \( P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} \)
- Understanding Mutually Exclusive Events: As in the problem, mutually exclusive events mean if event S occurs, event T cannot occur and vice versa. This simplifies how probabilities are worked out, because their joint probability is zero.
Common Denominator
When dealing with fractions, a Common Denominator is essential for adding or subtracting them. Particular to this exercise, to add the probabilities \( \frac{3}{5} \) and \( \frac{1}{3} \), we need to find a common denominator. The criteria for choosing a common denominator are:
- Definition: A common denominator is a shared multiple of the denominators of two or more fractions.
- Finding a Common Denominator: For fractions \( \frac{3}{5} \) and \( \frac{1}{3} \), a shared multiple is 15. Convert each fraction to an equivalent fraction with the common denominator.
- Conversion Steps: \( \frac{3}{5} = \frac{9}{15} \) and \( \frac{1}{3} = \frac{5}{15} \).
Simplification of Fractions
After converting fractions to have a common denominator, the next step is the Simplification of Fractions. Once the fractions share a common denominator, they can be added and simplified as necessary:
- Add the Fractions: Once the denominators are the same, you can simply add the numerators. For our problem, \( \frac{9}{15} + \frac{5}{15} = \frac{14}{15} \).
- Check for Simplification: Simplifying a fraction involves reducing it to the smallest possible terms, by dividing both the numerator and the denominator by their greatest common divisor (GCD). Here, \( \frac{14}{15} \) is already in its simplest form.
Other exercises in this chapter
Problem 14
Sketch the asymptotes and the graph of each equation. \(y=\frac{1}{x}-3\)
View solution Problem 14
Suppose that \(x\) and \(y\) vary inversely. Write a function that models each inverse variation and find \(y\) when \(x=10 .\) $$ x=20 \text { when } y=-4 $$
View solution Problem 15
Solve each equation. Check each solution. $$ \frac{11}{3 x}-\frac{1}{3}=\frac{-4}{x^{2}} $$
View solution Problem 15
Divide. State any restrictions on the variables. $$ \frac{6 x+6 y}{x-y} \div \frac{18}{5 x-5 y} $$
View solution