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Let \(A\) and \(B\) be events having positive probability. State whether each of the following statements is (i) necessarily true, (ii) necessarily false, or (iii) possibly true. (a) If \(A\) and \(B\) are mutually exclusive, then they are independent. (b) If \(A\) and \(B\) are independent, then they are mutually exclusive. (c) \(P(A)=P(B)=.6\), and \(A\) and \(B\) are mutually exclusive. (d) \(P(A)=P(B)=.6\), and \(A\) and \(B\) are independent.

Short Answer

Expert verified
(a) (ii) necessarily false (b) (iii) possibly true (c) (iii) possibly true (d) (i) necessarily true

Step by step solution

01

a. Mutually exclusive events and independence

If events A and B are mutually exclusive, this means that they cannot both occur at the same time. In other words, if A happens, then B cannot happen, and vice versa. Mathematically, this translates to \(P(A \cap B) = 0\). On the other hand, events A and B are independent if their joint probability equals the product of their individual probabilities, i.e., \(P(A \cap B) = P(A) \cdot P(B)\). Since both events have positive probabilities, their product cannot be equal to zero. Therefore, this statement is (ii) necessarily false.
02

b. Independent events and mutual exclusivity

If events A and B are independent, then their joint probability equals the product of their individual probabilities: \(P(A \cap B) = P(A) \cdot P(B)\). However, this does not necessarily mean that events A and B are mutually exclusive because their joint probability could be non-zero. In this case, this statement is (iii) possibly true, since there could be events that are both independent and mutually exclusive, but it's not guaranteed.
03

c. Mutually exclusive events with given probabilities

The given statement states that \(P(A) = P(B) = 0.6\), and events A and B are mutually exclusive. According to the definition of mutually exclusive events, \(P(A \cap B) = 0\). This statement is (iii) possibly true because there is no contradiction between the given probabilities of A and B and their mutual exclusivity.
04

d. Independent events with given probabilities

The given statement states that \(P(A) = P(B) = 0.6\), and events A and B are independent. For events to be independent, their joint probability should equal the product of their individual probabilities: \(P(A \cap B) = P(A) \cdot P(B)\). In this case, the joint probability would be \(0.6 \cdot 0.6 = 0.36\). Since the joint probability is not equal to zero or exceeding the individual probabilities, this statement is (i) necessarily true, because it satisfies the condition for independent events.

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Key Concepts

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

Mutually Exclusive Events
Understanding mutually exclusive events is critical in probability. If two events are mutually exclusive, it means they cannot occur at the same time. Picture drawing a card from a standard deck: drawing an ace of hearts (event A) and drawing a king of spades (event B) are mutually exclusive events, as you cannot draw both at the same instant.

Formally, for two events A and B that are mutually exclusive, we have the formula: \(P(A \cap B) = 0\). This means that the probability of both events occurring together (the intersection) is zero. In the context of our exercise, if A and B are mutually exclusive with positive probabilities, there's an inherent contradiction in saying they're also independent because by definition, independent events can occur together, they just don't influence each other's probability.
Independent Events
Independent events are fundamentally different from mutually exclusive events. When we say two events are independent, we're stating that the occurrence of one event does not affect the probability of the occurrence of the other. Real-life examples include flipping a coin and rolling a die simultaneously; whether you get heads or tails does not influence the roll's outcome.

Mathematically, independence is defined as: \(P(A \cap B) = P(A) \cdot P(B)\). Simply put, the probability that both A and B occur is the product of their individual probabilities. The provided problem shows that events being independent does not necessitate mutual exclusivity. Independence is about probabilities not influencing each other, while mutual exclusivity is about the impossibility of co-occurrence.
Joint Probability
When exploring joint probability, we deal with the likelihood of two events occurring together. In probability notation, we depict the joint probability of events A and B as \(P(A \cap B)\). Understanding joint probability is crucial when determining if events are independent or not. The calculation of joint probability differs based on whether the events are independent or not.

Calculating Joint Probability

For independent events, we use the product rule: \(P(A \cap B) = P(A) \cdot P(B)\). However, if the events are not independent, we must consider additional information, such as conditional probabilities, to find the joint probability accurately. Our aforementioned exercise uses joint probability to demonstrate the conditions of independence and mutual exclusivity.
Probability Theory
The backbone of our discussion on events and their relationships is probability theory. It is a branch of mathematics that measures the likelihood of events in terms of probabilities, ranging from 0 (impossible event) to 1 (certain event). Its principles govern random phenomena, such as rolling dice or market fluctuations.

Probability theory places a structure around the nature of randomness. It includes concepts like outcomes, events, and notably, the rules governing the behavior of probabilities—including the addition rule for mutually exclusive events and the multiplication rule for independent events. This theoretical framework enables us to systemically approach problems like the ones presented in the exercise. It illustrates that understanding the foundational definitions and rules of probability can guide us in discerning the truth of statements about events and the interplays between their probabilities.

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Most popular questions from this chapter

Consider the following game. A deck of cards is shuffled and its cards are turned face up one at a time. At any time you can elect to say "next," and if the next card is the ace of spades, then you win, and if not, then you lose. Of course, if the ace of spades appears before you say "next," then you lose. Also, if there is only one card remaining, the ace of spades hasn't yet appeared, and you have never said "next," then you are a winner (since you will say "next"). Argue that no matter what strategy you employ for deciding when to say "next," your probability of winning is \(\frac{1}{52}\).

A friend randomly chooses two cards, without replacement, from an ordinary deck of 52 playing cards. In each of the following situations, determine the conditional probability that both cards are aces. (a) You ask your friend if one of the cards is the ace of spades and your friend answers in the affirmative. (b) You ask your friend if the first card selected is an ace and your friend answers in the affirmative. (c) You ask your friend if the second card selected is an ace and your friend answers in the affirmative. (d) You ask your friend if either of the cards selected is an ace and your friend answers in the affirmative.

The probability of getting a head on a single toss of a coin is \(p\). Consider that \(A\) starts and continues to flip the coin until a tail shows up, at which point \(B\) starts flipping. Then \(B\) continues to flip until a tail comes up, at which point \(A\) takes over, and so on. Let \(P_{n, m}\) denote the probability that \(A\) accumulates a total of \(n\) heads before \(B\) accumulates \(m\). Show that $$ P_{n, m}=p P_{n-1, m}+(1-p)\left(1-P_{m, n}\right) $$

An engineering system consisting of \(n\) components is said to be a \(k\)-out- of\(n\) system \((k \leq n)\) if the system functions if and only if at least \(k\) of the \(n\) components function. Suppose that all components function independently of each other. (a) If the \(i\) th component functions with probability \(P_{t}, i=1,2,3,4\), compute the probability that a 2-out-of-4 system functions. (b) Repeat part (a) for a 3-out-of-5 system. (c) Repeat for a \(k\)-out-of- \(n\) system when all the \(P_{i}\) equal \(p\) (that is, \(P_{i}=p\), \(i=1,2, \ldots, n)\)

Consider the gambler's ruin problem with the exception that \(A\) and \(B\) agree to play no more than \(n\) games. Let \(P_{n, i}\) denote the probability that \(A\) winds up with all the money when \(A\) starts with \(i\) and \(B\) with \(N-i\). Derive an equation for \(P_{n, i}\) in terms of \(P_{n-1, i+1}\) and \(P_{n-1, i-1}\) and compute \(P_{7,3}\), \(N=5\)

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