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If \(A\) and \(B\) are mutually exclusive events and \(P(A)=0.2\) and \(P(B)=0.5\), find (a) \(P(A \cup B)\) (b) \(P(\bar{A})\) (c) \(P(\bar{A} \cap B)\)

Short Answer

Expert verified
(a) 0.7, (b) 0.8, (c) 0.5.

Step by step solution

01

Understanding Mutually Exclusive Events

Mutually exclusive events cannot occur at the same time. This means that if event A occurs, event B cannot occur and vice versa. For mutually exclusive events \(A\) and \(B\), \(P(A \cap B) = 0\).
02

Calculate \(P(A \cup B)\)

The probability of the union of two mutually exclusive events is the sum of their individual probabilities. Use the formula:\[ P(A \cup B) = P(A) + P(B) \]Given that \(P(A) = 0.2\) and \(P(B) = 0.5\), we have:\[ P(A \cup B) = 0.2 + 0.5 = 0.7 \]
03

Calculate \(P(\bar{A})\)

The complement of \(A\), denoted \(\bar{A}\), represents the event that \(A\) does not occur. Use the formula for the probability of the complement:\[ P(\bar{A}) = 1 - P(A) \]Given that \(P(A) = 0.2\), we find:\[ P(\bar{A}) = 1 - 0.2 = 0.8 \]
04

Calculate \(P(\bar{A} \cap B)\)

The event \(\bar{A} \cap B\) represents the occurrence of \(B\) without \(A\). Since \(B\) cannot occur with \(A\) due to mutual exclusivity, the probability of this event is just \(P(B)\). Hence:\[ P(\bar{A} \cap B) = P(B) = 0.5 \]

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Key Concepts

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

Mutually Exclusive Events
Mutually exclusive events are a fundamental concept in probability theory. These events cannot happen simultaneously, meaning the occurrence of one event excludes the occurrence of the other. For instance, if you flip a coin, the result could be either "heads" or "tails," but not both. Thus, these outcomes are mutually exclusive.

In probability terms, when events are mutually exclusive, the intersection of these events equals zero, expressed as:
  • \( P(A \cap B) = 0 \)
When given two mutually exclusive events \(A\) and \(B\), it is useful to identify them to simplify probability calculations. If we want to find the probability of either of these events occurring, we use their union. The probability of their union is simply the sum of their individual probabilities, as there is no overlap between them. This can be shown as:
  • \( P(A \cup B) = P(A) + P(B) \)
Complementary Events
Understanding complementary events is essential for mastering basic probability theory. A complementary event is essentially the opposite of a given event. If event \(A\) occurs, the complement of \(A\), denoted as \(\bar{A}\), represents all other possible outcomes where \(A\) does not occur.

One of the fundamental principles is that the probabilities of an event and its complement always add up to 1. This is written as:
  • \( P(A) + P(\bar{A}) = 1 \)
Therefore, to calculate the probability of the complement, simply subtract the probability of the event from 1, as seen in the equation:
  • \( P(\bar{A}) = 1 - P(A) \)
For example, if the probability of rain tomorrow is 0.7, the probability of no rain (the complement) is 0.3. This relationship simplifies solving probability problems by focus on one outcome possibility.
Probability Union and Intersection
Grasping the concepts of probability union and intersection helps in evaluating scenarios with multiple events. Let's break down these terms for better clarity.

The **union** of two events, \(A\) and \(B\) (written as \(A \cup B\)), represents the probability that at least one of the events occurs. For mutually exclusive events, which we've discussed, this is simply the sum of their individual probabilities but more generally, it is calculated by:
  • \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \)
This formula adjusts for any overlap that occurs in the intersection of these events.

**Intersection,** on the other hand, considers the scenario where both events occur simultaneously. It is denoted \(A \cap B\) and can sometimes be zero, especially in mutually exclusive events, as there is no overlap.

When faced problems involving these concepts, identifying whether events are mutually exclusive greatly influences which formulas to use, guiding the solution path. Understanding these principles underpins more complex probability calculations and applications.

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Most popular questions from this chapter

Ten thousand numbers are to be added, each rounded to the sixth decimal place. Assuming that the errors arising from rounding the numbers are mutually independent and uniformly distributed on \(\left(-0.5 \times 10^{-6},+0.5 \times 10^{-6}\right)\), find the limits in which the total error will lie with probability \(95 \%\).

The diameter of ball-bearings produced by a machine is a random variable having a normal distribution with mean \(6.00 \mathrm{~mm}\) and standard deviation \(0.02 \mathrm{~mm}\). If the diameter tolerance is \(\pm 1 \%\), find the proportion of ball-bearings produced that are out of tolerance. After several years' use, machine wear has the effect of increasing the standard deviation, although the mean diameter remains constant. The manufacturer decides to replace the machine when \(2 \%\) of its output is out of tolerance. What is the standard deviation when this happens?

The function. $$ \Gamma(\alpha)=\int_{0}^{\infty} y^{a-1} \mathrm{e}^{-y} \mathrm{~d} y \quad(\alpha>0) $$ is known as the gamma function, and the probability density function, $$ f_{X}(x)=\left\\{\begin{array}{cl} {[\Gamma(\alpha)]^{-1} \lambda^{\alpha} x^{\alpha-1} \mathrm{e}^{-\lambda x}} & (x>0) \\ 0 & \text { (otherwise) } \end{array}\right. $$ defines the gamma distribution. Prove that (a) \(\mu_{X}=\alpha / \lambda\) (b) \(\sigma_{X}^{2}=\alpha / \lambda^{2}\)

The distribution of downtime \(T\) for breakdowns of a computer system is given by $$ f_{T}(t)=\left\\{\begin{array}{cl} a^{2} t \mathrm{e}^{-a t} & (t>0) \\ 0 & \text { (otherwise) } \end{array}\right. $$ where \(a\) is a positive constant. The cost of downtime derived from the disruption resulting from breakdowns rises exponentially with \(T\) : $$ \text { cost factor }=h(T)=\mathrm{e}^{b T} $$ Show that the expected cost factor for downtime is \([a /(a-b)]^{2}\), provided that \(a>b\)

Part of an electrical circuit consists of three elements \(K, L\) and \(M\) in series. Probabilities of failure for elements \(K\) and \(M\) during operating time \(t\) are \(0.1\) and \(0.2\) respectively. Element \(L\) itself consists of three sub-elements \(L_{1}, L_{2}\) and \(L_{3}\) in parallel, with failure probabilities \(0.4,0.7\) and \(0.5\) respectively, during the same operating time \(t\). Find the probability of failure of the circuit during time \(t\), assuming that all failures of elements are independent.

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