Problem 72
Question
If \(\bar{E}\) and \(\bar{F}\) are the complementary events of events \(E\) and \(F\) respectively and if \(0
Step-by-Step Solution
Verified Answer
Options (A) and (D) are true.
1Step 1: Understanding Complementary Events
Complementary events are events that cover all possible outcomes of an experiment. For instance, if \(E\) is an event, then \(\bar{E}\), the complement of \(E\), is an event that occurs when \(E\) does not occur. Thus, the probabilities associated with complementary events are \(P(E) + P(\bar{E}) = 1\). This holds true for any event \(F\) as well.
2Step 2: Analyze Each Option
We need to see which of the event combinations involving \(E\), \(\bar{E}\), \(F\), \(\bar{F}\) is always equal to 1. Recall: \(P(A \mid B)\) is the probability of \(A\) given \(B\).
3Step 3: Examine Option A
For option (A), we have \(P(E \mid F) + P(\bar{E} \mid F) = 1\). Given that \(\bar{E}\) is the complementary of \(E\), within set \(F\), these are complementary probabilities. Thus, since all outcomes are covered, the sum is indeed 1.
4Step 4: Evaluate Option B
For option (B), \(P(E \mid F) + P(E \mid \bar{F})\) does not necessarily add up to 1. \(P(E \mid F)\) and \(P(E \mid \bar{F})\) refer to different events (conditional probabilities of \(E\) given \(F\) and \(\bar{F}\), respectively), so they don't complement each other within the same sample space or event.
5Step 5: Evaluate Option C
For option (C), we have \(P(\bar{E} \mid F) + P(E \mid \bar{F})\). These represent different events too: \(E\) given \(\bar{F}\) and \(\bar{E}\) given \(F\), they do not relate as complements within the same conditional context so they sum to a value that is not constrained to be 1.
6Step 6: Evaluate Option D
In option (D), \(P(E \mid \bar{F}) + P(\bar{E} \mid \bar{F})\) equates to 1 because within the conditional context of \(\bar{F}\), \(E\) and \(\bar{E}\) are complementary events and their probabilities must sum to 1.
7Step 7: Identify the Correct Answer
The possibilities out of the four options that are structured to show the completeness of complementary probabilities sum to 1 are (A) and (D). These options respect the complementary principle within the context of their respective conditioning sets.
Key Concepts
Conditional ProbabilityComplementary EventsEvent Probabilities
Conditional Probability
Conditional probability is a fundamental concept in probability theory. It represents the likelihood of an event occurring given that another event has already occurred. This is typically described using the notation \(P(A \mid B)\), which reads as "the probability of \(A\) given \(B\)."
Understanding conditional probability is crucial because it allows for a more nuanced view of probability, one that considers the influence of certain conditions or pre-existing events on the probability of another event. Conditional probability is calculated using the formula:
In many real-world situations, calculating the probability of an event influenced by another event helps in decision-making, risk assessment, and predictive modeling. It narrows down possible outcomes to a subset where the condition holds true, providing a clearer picture of what might happen.
Understanding conditional probability is crucial because it allows for a more nuanced view of probability, one that considers the influence of certain conditions or pre-existing events on the probability of another event. Conditional probability is calculated using the formula:
- \(P(A \mid B) = \frac{P(A \cap B)}{P(B)}\) if \(P(B) > 0\)
In many real-world situations, calculating the probability of an event influenced by another event helps in decision-making, risk assessment, and predictive modeling. It narrows down possible outcomes to a subset where the condition holds true, providing a clearer picture of what might happen.
Complementary Events
Complementary events are pairs of outcomes that make up the entire sample space of an event. When one of these events occurs, the other does not, and vice versa. Mathematically, if \(E\) is an event, then \(\bar{E}\) is its complement.
An important property of complementary events is that their probabilities add up to 1:
Understanding complementary events is essential because they allow us to assess the likelihood of an event by simply knowing what is not the event. This relationship can also help simplify complex probability problems by using the complement rule, which states:
An important property of complementary events is that their probabilities add up to 1:
- \(P(E) + P(\bar{E}) = 1\)
Understanding complementary events is essential because they allow us to assess the likelihood of an event by simply knowing what is not the event. This relationship can also help simplify complex probability problems by using the complement rule, which states:
- \(P(\bar{E}) = 1 - P(E)\)
Event Probabilities
Event probabilities are the backbone of probability theory. Each event in an experiment or sample space is assigned a probability, a number between 0 and 1, reflecting its likelihood of occurring.
Probabilities can be determined using different methods:
Probabilities can be determined using different methods:
- Theoretical probability, which is based on the expectation due to symmetry or logical basis. For example, the probability of rolling a 3 on a fair die is \(\frac{1}{6}\).
- Empirical probability, which is based on experiment or historical data, calculated as \(\frac{\text{Number of successful outcomes}}{\text{Total number of trials}}\).
- Subjective probability, which derives from personal judgment or estimation.
- Non-negativity: \(P(E) \geq 0\)
- Normalization: \(P(S) = 1\) for the entire sample space \(S\)
- Additivity: For mutually exclusive events \(A\) and \(B\), \(P(A \cup B) = P(A) + P(B)\)
Other exercises in this chapter
Problem 70
A bag contains \(n+1\) coins. It is known that one of these coins shows heads on both sides, whereas the other coins are fair. One coin is selected at random an
View solution Problem 71
The probabilites of three events \(A, B\) and \(C\) are \(P(A)=0.6, P(B)=0.4\) and \(P(C)=0.5 .\) If \(P(A \cup B)\) \(=0.8, P(A \cap C)=0.3, P(A \cap B \cap C)
View solution Problem 73
The probability that certain electronic component fails when first used is \(0.10\). If it does not fail immediately, the probability that it lasts for one year
View solution Problem 74
Four tickets marked \(00,01,10\) and 11 , respectively, are placed in a bag. A ticket is drawn at random five times, being replaced each time. The probability t
View solution