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Consider the following addition rule to find the probability of the union of two events \(A\) and \(B\) : $$P(A \text { or } B)=P(A)+P(B)-P(A \text { and } B)$$ When and why is the term \(P(A\) and \(B)\) subtracted from the sum of \(P(A)\) and \(P(B)\) ? Give one example where you might use this formula.

Short Answer

Expert verified
The term \(P(A \text{ and } B)\) is subtracted from \(P(A) + P(B)\) to remove the overlap that arises from adding the probabilities of \(A\) and \(B\), as the cases where both \(A\) and \(B\) happen are counted twice. This formula is useful when you want to find the probability of either of two events happening that are not mutually exclusive. For example, when calculating the probability of drawing a red card or a queen from a deck of cards.

Step by step solution

01

Explanation of Deduction of \(P(A \text{ and } B)\)

The term \(P(A \text{ and } B)\) is subtracted to remove the overlap, or the joint probability of \(A\) and \(B\). This is because when adding the probabilities \(P(A)\) and \(P(B)\), the cases where both \(A\) and \(B\) happen are counted twice. So, to account for this double counting, the probability of the occurrences of both \(A\) and \(B\) is subtracted.
02

When to Use This Formula

This formula is used when it is needed to find the probability of either of two events happening. It's particularly necessary when the two events aren't mutually exclusive - meaning they can both occur at the same time.
03

Example

Imagine there is a deck of cards (52 cards) and want to find the probability of drawing a red card or drawing a queen. First, find the probability of drawing a red card \(P(A) = \frac{26}{52} = 0.5\). Next, calculate the probability of drawing a queen card \(P(B) = \frac{4}{52} = 0.077\). Since there are two red queen cards in the deck, the probability of drawing a red queen card, or \(P(A \text{ and } B) = \frac{2}{52} = 0.038\). Apply those numbers to the rule: \(P(A \text { or } B) = P(A) + P(B) - P(A \text { and } B) = 0.5 + 0.077 - 0.038 = 0.539. Hence, the probability of drawing a red card or a queen card is 0.539.

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Key Concepts

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

Addition Rule
In probability, the addition rule is a fundamental concept. It helps us determine the probability of the occurrence of at least one of two events. The formula for the addition rule is: \[P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)\]. This formula is extremely useful because it accounts for scenarios where the two events might somehow overlap. Whenever two events A and B are considered, adding their individual probabilities directly will include the overlapping section twice. Therefore, to correct this and ensure accurate results, we subtract the portion where both events occur simultaneously.

It simplifies finding the combined probability of complex scenarios. Even for someone new to probability, the rule's ease of use and clear logic make it an invaluable tool. Understanding when and how to apply it is key to solving probability problems efficiently.
Probability of the Union
The probability of the union of two events refers to the likelihood that either event A or event B, or both, will occur. This is what the addition rule seeks to calculate. When dealing with the union of events, we are essentially interested in any outcome that falls within either event's possible results. The formula to determine the union is: \[P(A \cup B) = P(A) + P(B) - P(A \cap B)\].

Here, the symbol \(A \cup B\) represents the union of sets, and \(A \cap B\) signifies the intersection. The intersection is what we subtract since it represents the outcomes common to both events A and B that we might have counted more than once. This consideration is why this formula is especially useful with non-mutually exclusive events. It ensures that no probability is overstated, providing precise and accurate results when calculating the outcomes.
Non-Mutually Exclusive Events
Non-mutually exclusive events are events that can occur at the same time. This is opposed to mutually exclusive events, which cannot happen simultaneously. The addition rule stands out in such cases, where events overlap, and without careful consideration, might lead to overestimated probabilities if added directly. For example, let's revisit the exercise: Drawing a red card or a queen from a deck of cards. These events are non-mutually exclusive because you can draw a red queen, which is both a red card and a queen. This overlap must be subtracted (the joint probability) so that we only count it once, making that scenario an ideal candidate for the addition rule.

Understanding this concept helps us manage outcomes comprehensively and avoids common pitfalls in probability calculations.
Joint Probability
Joint probability is the probability of two or more events occurring simultaneously. When we talk about the joint probability of events A and B, we denote it by \(P(A \cap B)\). It's vital because it represents the overlap between events in the addition rule formula. In the provided exercise, the joint probability \(P(A \text{ and } B)\) is determined as the probability of drawing a red queen. Such dual events signify the need for careful computation to ensure they are neither overlooked nor overcounted.

Joint probability becomes an integral part of understanding event relationships, particularly when those events aren't mutually exclusive. As such, grasping joint probability is essential in working through probabilities of combined, simultaneous or overlapping scenarios, ensuring clarity in broader probability assessments.

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Most popular questions from this chapter

Consider the following games with two dice. a. A gambler is going to roll a die four times. If he rolls at least one 6, you must pay him \(\$ 5 .\) If he fails to roll a 6 in four tries, he will pay you \(\$ 5\). Find the probability that you must pay the gambler. Assume that there is no cheating. b. The same gambler offers to let you roll a pair of dice 24 times. If you roll at least one double 6 , he will pay you \(\$ 10\). If you fail to roll a double 6 in 24 tries, you will pay him \(\$ 10\). The gambler says that you have a better chance of winning because your probability of success on each of the 24 rolls is \(1 / 36\) and you have 24 chances. Thus, he says, your probability of winning \(\$ 10\) is \(24(1 / 36)=2 / 3\). Do you agree with this analysis? If so, indicate why. If not, point out the fallacy in his argument, and then find the correct probability that you will win.

A trimotor plane has three engines-a central engine and an engine on each wing. The plane will crash only if the central engine fails and at least one of the two wing engines fails. The probability of failure during any given flight is \(.005\) for the central engine and \(.008\) for each of the wing engines. Assuming that the three engines operate independently, what is the probability that the plane will crash during a flight?

A box contains 10 red marbles and 10 green marbles. a. Sampling at random from the box five times with replacement, you have drawn a red marble all five times. What is the probability of drawing a red marble the sixth time? b. Sampling at random from the box five times without replacement, you have drawn a red marble all five times. Without replacing any of the marbles, what is the probability of drawing a red marble the sixth time? c. You have tossed a fair coin five times and have obtained heads all five times. A friend argues that according to the law of averages, a tail is due to occur and, hence, the probability of obtaining a head on the sixth toss is less than \(.50 .\) Is he right? Is coin tossing mathematically equivalent to the procedure mentioned in part a or the procedure mentioned in part b? Explain.

The probability that a student graduating from Suburban State University has student loans to pay off after graduation is \(.60\). If two students are randomly selected from this university, what is the probability that neither of them has student loans to pay off after graduation?

A consumer agency randomly selected 1700 flights for two major airlines, \(\mathrm{A}\) and \(\mathrm{B}\). The following table gives the two-way classification of these flights based on airline and arrival time. Note that "less than 30 minutes late" includes flights that arrived early or on time. $$\begin{array}{cccc} \hline & \begin{array}{c} \text { Less Than 30 } \\ \text { Minutes Late } \end{array} & \begin{array}{c} \mathbf{3 0} \text { Minutes to } \\ \text { 1 Hour Late } \end{array} & \begin{array}{c} \text { More Than } \\ \text { 1 Hour Late } \end{array} \\ \hline \text { Airline A } & 429 & 390 & 92 \\ \text { Airline B } & 393 & 316 & 80 \\ \hline \end{array}$$ a. Suppose one flight is selected at random from these 1700 flights. Find the following probabilities. i. \(P(\) more than 1 hour late and airline \(\mathrm{A}\) ) ii. \(P(\) airline \(\mathrm{B}\) and less than 30 minutes late) b. Find the joint probability of events " 30 minutes to 1 hour late" and "more than 1 hour late." Is this probability zero? Explain why or why not.

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