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Brain Teasers Assume \(A\) and \(B\) are events such that \(0

Short Answer

Expert verified
True, because \(P(A | B) = \frac{P(A \cap B)}{P(B)}\), dividing by \(P(B)<1\) increases the value.

Step by step solution

01

Understanding Conditional Probability

The conditional probability \(P(A | B)\) is defined as the probability that event \(A\) occurs given that event \(B\) has already occurred. Mathematically, it is expressed as \(P(A | B) = \frac{P(A \cap B)}{P(B)}\).
02

Probability Ranges and Multiplication Rule

The probability of an event occurring is always between 0 and 1. Therefore, \(P(A \cap B)\), which is the probability of both events \(A\) and \(B\) occurring together, also lies between 0 and 1. The relationship \(P(A | B) = \frac{P(A \cap B)}{P(B)}\) implies that \(P(A | B)\) is influenced by \(P(B)\).
03

Inequality Evaluation

Given that \(0 < P(B) < 1\), \(P(A \cap B)\) is divided by a number between 0 and 1 to obtain \(P(A | B)\). This division increases \(P(A \cap B)\) if \(P(B)\) is less than 1 (which it is by assumption), thereby implying \(P(A | B) \geq P(A \cap B)\).
04

Conclusion

Therefore, the statement \(P(A | B) \geq P(A \cap B)\) is true. This conclusion follows directly from the properties of probabilities and conditional probability.

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Key Concepts

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

Probability Rules
Probability rules are a set of guidelines that help you calculate the likelihood of events occurring. They're fundamental to understanding and solving many probability problems. One of the basic rules you need to know is that the probability of any event is always a number between 0 and 1. This means an event cannot occur less than 0% of the time or more than 100% of the time.

These rules also tell us how probabilities work with different events. For instance, if you're looking at the probability of two events occurring together, like event A and event B, you use the rule of multiplication for independent events. That is, if A and B are independent, you calculate the joint probability as:
  • \(P(A \text{ and } B) = P(A) \times P(B)\).

However, when events are not independent, which often happens in real-world situations, the multiplication rule adjusts by incorporating conditional probability. This rule takes into account how the occurrence of one event can affect the probability of another. The formula for conditional probability is:
  • \(P(A | B) = \frac{P(A \cap B)}{P(B)}\).
Understanding these rules will allow you to tackle more complex problems and dissect them into simpler parts.
Independent Events
Events are considered independent when the occurrence of one event does not influence the occurrence of another. The world of probability often involves determining whether events are independent or dependent. If they are independent, you can use the simple multiplication rule to find joint probabilities, knowing that these events do not affect each other.

Think of flipping a coin multiple times. Each flip is independent of the others. The outcome of one flip doesn't affect the others. The probability of getting heads twice in a row is determined by multiplying the probability of heads on the first flip by the probability on the second flip. So,
  • \(P(\text{heads on the first flip}) \times P(\text{heads on the second flip}) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}\).
However, many situations involve dependent events. Unlike independent events, calculating joint probabilities for dependent events requires using conditional probabilities to adjust for one event's influence on the other. For instance, if you're drawing cards from a deck without replacement, the probability of future draws is influenced by the cards that have already been removed.
Probability Range
The concept of probability range is central to understanding probabilities. It defines that any probability must lie within the range of 0 and 1, inclusive. This range tells us that an event's chance of occurring is never lower than 0% (impossible) and never higher than 100% (certain).

This range also helps in interpreting probabilities, allowing us to understand real-world scenarios better. For example, if there is a probability of 0.7 for an event, we interpret this as a 70% chance, suggesting a fairly likely occurrence. On the other hand, a probability of 0.05 indicates a 5% chance, which is relatively unlikely.

Understanding this range is useful when observing conditional probabilities and certain inequalities in probability. Importantly, when given a probability like \(P(A | B)\), we know it results from dividing \(P(A \cap B)\) by \(P(B)\), which inherently constrains \(P(A | B)\) to the range 0 to 1. This consistent framework aids in evaluating more advanced probability statements, ensuring they're accurate and make logical sense.

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Most popular questions from this chapter

Survey: Customer Loyalty Are customers more loyal in the east or in the west? The following table is based on information from Trends in the United States, published by the Food Marketing Institute, Washington, D.C. The columns represent length of customer loyalty (in years) at a primary supermarket. The rows represent regions of the United States. $$\begin{array}{lccccccc} \hline & \begin{array}{c} \text { Less Than } \\ \text { 1 Year } \end{array} & \begin{array}{c} 1-2 \\ \text { Years } \end{array} & \begin{array}{c} 3-4 \\ \text { Years } \end{array} & \begin{array}{c} 5-9 \\ \text { Years } \end{array} & \begin{array}{c} 10-14 \\ \text { Years } \end{array} & \begin{array}{c} 15 \text { or More } \\ \text { Years } \end{array} & \begin{array}{c} \text { Row } \\ \text { Total } \end{array} \\ \hline \text { East } & 32 & 54 & 59 & 112 & 77 & 118 & 452 \\ \text { Midwest } & 31 & 68 & 68 & 120 & 63 & 173 & 523 \\ \text { South } & 53 & 92 & 93 & 158 & 106 & 158 & 660 \\ \text { West } & 41 & 56 & 67 & 78 & 45 & 86 & 373 \\ \text { Column Total } & 157 & 270 & 287 & 468 & 291 & 535 & 2008 \\ \hline \end{array}$$ What is the probability that a customer chosen at random (a) has been loyal 10 to 14 years? (b) has been loyal 10 to 14 years, given that he or she is from the east? (c) has been loyal at least 10 years? (d) has been loyal \(a t\) least 10 years, given that he or she is from the west? (e) is from the west, given that he or she has been loyal less than I year? (f) is from the south, given that he or she has been loyal less than 1 year? (g) has been loyal I or more years, given that he or she is from the east? (h) has been loyal I or more years, given that he or she is from the west? (i) Are the events "from the east" and "loyal 15 or more years" independent? Explain.

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.

Brain Teasers Assume \(A\) and \(B\) are events such that \(0

Basic Computations: Rules of Probability Given \(P(A)=0.2, P(B)=0.5\) \(P(A | B)=0.3.\) (a) Compute \(P(A \text { and } B).\) (b) Compute \(P(A \text { or } B).\)

You toss a pair of dice. (a) Determine the number of possible pairs of outcomes. (Recall that there are six possible outcomes for each die.) (b) There are three even numbers on each die. How many outcomes are possible with even numbers appearing on each die? (c) Probability extension: What is the probability that both dice will show an even number?

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