Problem 14
Question
\(\boldsymbol{S}\) and \(\boldsymbol{T}\) are mutually exclusive events. Find \(\boldsymbol{P}(\boldsymbol{S} \text { or } \boldsymbol{T})\) $$ P(S)=\frac{5}{8}, P(T)=\frac{1}{8} $$
Step-by-Step Solution
Verified Answer
The probability of either S or T occurring is \(\frac{3}{4}\) or 75%
1Step 1: Define the Problem
The first step is understand the problem. Mutually exclusive events in probability theory are two events that cannot both occur. An event is said to be either 'S' or 'T'. The problem wants to find the probability that event 'S' or event 'T' will happen when it is known that 'S' and 'T' cannot both occur at the same time, and when the probability of each event occurring are: \( P(S) = \frac{5}{8} \) and \( P(T) = \frac{1}{8} \).
2Step 2: Apply the Formula for Probability of Mutually Exclusive Events
We need to use the formula for the probability of either of two mutually exclusive events occurring, that is: \(P(S \text{ or } T) = P(S) + P(T)\)
3Step 3: Substitute Values into the Formula
The probabilities given in the problem are \(P(S) = \frac{5}{8}\) and \(P(T) = \frac{1}{8}\). Substitute these values into the formula to find the probability of either S or T occurring. Thus, \(P(S \text{ or } T) = \frac{5}{8} + \frac{1}{8}\)
4Step 4: Calculate the Result
Now compute the result, \(P(S \text{ or } T) = \frac{6}{8}\)
5Step 5: Simplify the Result
Simplify the result to its lowest terms to get the answer. \(P(S \text{ or } T) = \frac{3}{4}\)
6Step 6: Comprehending the Result
The result, \(P(S \text{ or } T) = \frac{3}{4}\), means that there is a 75% chance for either event 'S' or event 'T' to occur.
Key Concepts
Mutually Exclusive EventsProbability TheoryAddition Rule for Probability
Mutually Exclusive Events
In probability theory, mutually exclusive events are situations where two events cannot happen at the same time. This means that if one event occurs, the other cannot. For example, when flipping a coin, landing on heads and tails are mutually exclusive events; the coin can only land on one side at a time. Understanding this concept helps when calculating probabilities because it simplifies how we find the likelihood of either event happening. Knowing that events are mutually exclusive allows us to use specific rules in probability theory, like the addition rule, to determine the overall probability of one event or the other happening.
Probability Theory
Probability theory is a branch of mathematics that deals with the likelihood of events occurring. It provides a framework for predicting the chances of different outcomes. This theory applies to various fields such as statistics, gambling, science, and engineering.
The basic unit in probability is an event, which is any specific outcome or group of outcomes. For example, rolling a dice and getting a five is an event. Probability assignment to an event ranges from zero (impossibility) to one (certainty). Understanding basic probability concepts like independent events, dependent events, and mutually exclusive events makes it easier to calculate and reason about complex scenarios.
The basic unit in probability is an event, which is any specific outcome or group of outcomes. For example, rolling a dice and getting a five is an event. Probability assignment to an event ranges from zero (impossibility) to one (certainty). Understanding basic probability concepts like independent events, dependent events, and mutually exclusive events makes it easier to calculate and reason about complex scenarios.
Addition Rule for Probability
The addition rule for probability is particularly useful when dealing with mutually exclusive events. This rule states that the probability of either of two mutually exclusive events occurring is simply the sum of their individual probabilities.
The formula is: \[ P(A \text{ or } B) = P(A) + P(B) \] when \( A \) and \( B \) are mutually exclusive events. For instance, in the exercise, since events \( S \) and \( T \) cannot occur simultaneously, calculating \( P(S \text{ or } T) \) involves adding \( P(S) \) and \( P(T) \). Therefore: \[ P(S \text{ or } T) = \frac{5}{8} + \frac{1}{8} = \frac{6}{8} = \frac{3}{4} \] This example illustrates how straightforward the calculation becomes with mutually exclusive events, providing a clear pathway to the solution.
The formula is: \[ P(A \text{ or } B) = P(A) + P(B) \] when \( A \) and \( B \) are mutually exclusive events. For instance, in the exercise, since events \( S \) and \( T \) cannot occur simultaneously, calculating \( P(S \text{ or } T) \) involves adding \( P(S) \) and \( P(T) \). Therefore: \[ P(S \text{ or } T) = \frac{5}{8} + \frac{1}{8} = \frac{6}{8} = \frac{3}{4} \] This example illustrates how straightforward the calculation becomes with mutually exclusive events, providing a clear pathway to the solution.
Other exercises in this chapter
Problem 13
The weight \(P\) in pounds that a beam can safely carry is inversely proportional to the distance \(D\) in feet between the supports of the beam. For a certain
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Divide. State any restrictions on the variables. $$ \frac{3 x^{3}}{5 y^{2}} \div \frac{6 x^{5}}{5 y^{3}} $$
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