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Basic Computation: Addition Rule Given \(P(A)=0.3\) and \(P(B)=0.4:\) (a) If \(A\) and \(B\) are mutually exclusive events, compute \(P(A \text { or } B)\) (b) If \(P(A \text { and } B)=0.1,\) compute \(P(A \text { or } B)\)

Short Answer

Expert verified
(a) 0.7; (b) 0.6

Step by step solution

01

Understanding the Addition Rule for Probability

The Addition Rule for Probability states that for any two events A and B, the probability that either event occurs is given by:\[ P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B) \]It's essential to understand this formula as it helps compute the probability for either event occurring.
02

Part (a): Mutually Exclusive Events

For mutually exclusive events, the occurrence of one event excludes the possibility of the other. Therefore, \( P(A \text{ and } B) = 0 \). So the formula becomes:\[ P(A \text{ or } B) = P(A) + P(B) \]Substituting the given values:\[ P(A \text{ or } B) = 0.3 + 0.4 = 0.7 \]
03

Part (b): General Case with Intersection

For non-mutually exclusive events, we have to account for the probability of both events occurring. Using the formula:\[ P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B) \]Substituting the given values including the intersection:\[ P(A \text{ or } B) = 0.3 + 0.4 - 0.1 = 0.6 \]

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Key Concepts

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

Addition Rule
The Addition Rule in probability helps us find out the likelihood of either one or two events happening. Rather than guessing, using this rule lets us calculate it exactly. The rule formula is: \[ P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B) \] Here,
  • \( P(A) \) is the probability of event A happening.
  • \( P(B) \) is the probability of event B happening.
  • \( P(A \text{ and } B) \) is the probability that both events occur.
It's crucial to subtract \( P(A \text{ and } B) \) to avoid double counting the times both events happen. Without this subtraction, our calculations might end up flawed, leading to probabilities greater than 1. In the case where events are mutually exclusive, this intersection term becomes zero, simplifying our lives by neatly reducing the formula.
Mutually Exclusive Events
Mutually exclusive events are special. When two events are mutually exclusive, it means they cannot happen at the same time. Think of flipping a coin and landing heads or tails but not both. For these events, the overlap or common outcome, represented by \( P(A \text{ and } B) \), is actually zero. Because both can't happen together, we simplify the addition rule: \[ P(A \text{ or } B) = P(A) + P(B) \]This simple addition works because we don't worry about subtracting any common outcomes. Knowing whether events are mutually exclusive helps us approach problems with the right strategy, like choosing the right formula and avoiding overcomplicated calculations. It gives better clarity in determining outcomes, simplifying complex scenarios.
Intersection of Events
The intersection of events in probability refers to situations where two events happen simultaneously. This concept is key in understanding overlaps between events. In probability notation, this is expressed as \( P(A \text{ and } B) \). When calculating probabilities using the addition rule, intersecting events are crucial. The formula for intersecting events is: \[ P(A \text{ and } B) \] When events are not mutually exclusive, the probability of the intersection must be considered to accurately reflect both events happening at the same time.
  • If the intersection probability \( P(A \text{ and } B) \) is non-zero, both events can occur together.
  • If zero, then events might be mutually exclusive, and they never happen at the same time, simplifying our work.
Being aware of intersections aids in fine-tuning our addition rule application, avoiding any unnecessary complexity or error in probability calculations.

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Most popular questions from this chapter

Grading One professor grades homework by randomly choosing 5 out of 12 homework problems to grade. (a) How many different groups of 5 problems can be chosen from the 12 problems? (b) Probability Extension Jerry did only 5 problems of one assignment. What is the probability that the problems he did comprised the group that was selected to be graded? (c) Silvia did 7 problems. How many different groups of 5 did she complete? What is the probability that one of the groups of 5 she completed comprised the group selected to be graded?

Sometimes probability statements are expressed in terms of odds. The odds in favor of an event \(A\) are the ratio \(\frac{P(A)}{P(n o t A)}=\frac{P(A)}{P\left(A^{c}\right)}\) For instance, if \(P(A)=0.60,\) then \(P\left(A^{C}\right)=0.40\) and the odds in favor of \(A\) are $$ \frac{0.60}{0.40}=\frac{6}{4}=\frac{3}{2}, \text { written as } 3 \text { to } 2 \text { or } 3: 2 $$ (a) Show that if we are given the odds in favor of event \(A\) as \(n: m,\) the probability of event \(A\) is given by \(P(A)=\frac{n}{n+m} .\) Hint: Solve the equation $$ \frac{n}{m}=\frac{P(A)}{1-P(A)} \text { for } P(A) $$ (b) A telemarketing supervisor tells a new worker that the odds of making a sale on a single call are 2 to \(15 .\) What is the probability of a successful call? (c) A sports announcer says that the odds a basketball player will make a free throw shot are 3 to \(5 .\) What is the probability the player will make the shot?

Involve a standard deck of 52 playing cards. In such a deck of cards there are four suits of 13 cards each. The four suits are: hearts, diamonds, clubs, and spades. The 26 cards included in hearts and diamonds are red. The 26 cards included in clubs and spades are black. The 13 cards in each suit are: \(2,3,4,5,6,7,8,9,10,\) Jack, Queen, King, and Ace. This means there are four Aces, four Kings, four Queens, four \(10 \mathrm{s},\) etc., down to four \(2 \mathrm{s}\) in each deck. You draw two cards from a standard deck of 52 cards, but before you draw the second card, you put the first one back and reshuffle the deck. (a) Are the outcomes on the two cards independent? Why? (b) Find \(P(3 \text { on } 1 \text { st card and } 10 \text { on } 2\)nd). (c) Find \(P(10 \text { on } 1 \text { st card and } 3 \text { on } 2 \text { nd })\) (d) Find the probability of drawing a 10 and a 3 in either order.

A recent Harris Poll survey of 1010 U.S. adults selected at random showed that 627 consider the occupation of firefighter to have very great prestige. Estimate the probability (to the nearest hundredth) that a U.S. adult selected at random thinks the occupation of firefighter has very great prestige.

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