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Let \(A\) and \(B\) be two events such that \(P(\overline{A \cup B})=1 / 6\). \(P(A \cap B)=1 / 4\) and \(P(\bar{A})=1 / 4\) where \(\bar{A}\) stands for complment of event \(A\). Then events \(A\) and \(B\) are a. equally likely but not independent b. equally likely and mutually exclusive c. mutually exclusive and independent d. independent but not equally likely

Short Answer

Expert verified
The events are independent but not equally likely (option d).

Step by step solution

01

Calculate P(A ∪ B) using its complement

We are given \(P(\overline{A \cup B}) = \frac{1}{6}\). The relationship between an event and its complement is \(P(A \cup B) + P(\overline{A \cup B}) = 1\). Thus, we can find \(P(A \cup B)\) by rearranging this equation:\[P(A \cup B) = 1 - P(\overline{A \cup B}) = 1 - \frac{1}{6} = \frac{5}{6}\]
02

Apply the formula for conditional probability

We use the formula: \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\). Rearranging this, we can find \(P(A) + P(B)\) because \(P(A \cup B) = \frac{5}{6}\) and \(P(A \cap B) = \frac{1}{4}\):\[P(A) + P(B) = P(A \cup B) + P(A \cap B) = \frac{5}{6} + \frac{1}{4} = \frac{31}{12}\]
03

Determine P(A) from its complement

We are given that \(P(\bar{A}) = \frac{1}{4}\), so \(P(A) = 1 - P(\bar{A})\) gives us:\[P(A) = 1 - \frac{1}{4} = \frac{3}{4}\]
04

Calculate P(B) from known probabilities

Using the information from Step 2, and the result of \(P(A) = \frac{3}{4}\) from Step 3, we can find \(P(B)\) using \(P(A) + P(B) = \frac{31}{12}\):\[P(B) = \frac{31}{12} - \frac{3}{4} = \frac{31}{12} - \frac{9}{12} = \frac{22}{12} = \frac{11}{6}\]
05

Check if events are independent

Two events, \(A\) and \(B\), are independent if \(P(A \cap B) = P(A) \cdot P(B)\). We can check this:- The product \(P(A) \cdot P(B)\) would be \(\frac{3}{4} \cdot \frac{11}{6} = \frac{33}{24} = \frac{11}{8}\).- Since \(P(A \cap B) = \frac{1}{4}\), and \(\frac{1}{4} eq \frac{11}{8}\), \(A\) and \(B\) are not independent.
06

Check if events are equally likely

Events \(A\) and \(B\) are equally likely if \(P(A) = P(B)\). From Step 3, \(P(A) = \frac{3}{4}\), and from Step 4, \(P(B) = \frac{11}{6}\). Clearly, \(\frac{3}{4} eq \frac{11}{6}\), so \(A\) and \(B\) are not equally likely.

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Key Concepts

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

Event Independence
In probability theory, understanding whether two events are independent is crucial. Two events, \(A\) and \(B\), are independent if the occurrence of one event does not influence the probability of the other event occurring. Mathematically, this is represented by the equation \(P(A \cap B) = P(A) \cdot P(B)\). This means the probability of both events happening together should equal the product of their individual probabilities.

In our exercise, we have \(P(A \cap B) = \frac{1}{4}\). To verify independence, we calculate \(P(A) \cdot P(B)\), which equals \(\frac{11}{8}\). Since \(\frac{1}{4} eq \frac{11}{8}\), events \(A\) and \(B\) are not independent. This highlights how independence requires a very specific condition in their probabilities.
Equally Likely Events
When two events are equally likely, it means that they have the same probability of occurring. In simpler terms, they are as likely to happen as each other. To determine this, we compare the individual probabilities of events \(A\) and \(B\).

From the solution, we calculated that \(P(A) = \frac{3}{4}\) and \(P(B) = \frac{11}{6}\). Since \(\frac{3}{4} eq \frac{11}{6}\), they are not equally likely. This tells us that in situations where probabilities differ, one event is more probable than the other, which can affect planning or predictions based on these events.
Mutual Exclusivity
The concept of mutual exclusivity refers to two events that cannot happen at the same time. If events \(A\) and \(B\) are mutually exclusive, then \(P(A \cap B) = 0\) because they cannot both occur simultaneously.

In our exercise, we found \(P(A \cap B) = \frac{1}{4}\), meaning events \(A\) and \(B\) can both happen, hence they are not mutually exclusive. The concept of mutual exclusivity is important in scenarios where the occurrence of one event definitely prevents the occurrence of another, such as rolling a single die and getting either a 3 or a 5, but not both.
Conditional Probability
Conditional probability is the probability of an event occurring given that another event has already occurred. It is represented as \(P(A|B)\), which reads as "the probability of \(A\) given \(B\)". This depends on the relationship between \(A\) and \(B\) and is calculated using the formula \(P(A|B) = \frac{P(A \cap B)}{P(B)}\).

In the given problem, conditional probability could have provided additional insights if needed, although the specific focus was not on this concept. Understanding how one event affects the likelihood of another is useful in many real-world situations, such as patient health conditions, weather forecasts, or any scenario where one condition alters outcomes for another.

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Most popular questions from this chapter

Two players \(P\), and \(P_{2}\) are playing the final of a chess championship, which consists of a series of matches. Probability of \(P_{1}\) winning a match is \(2 / 3\) and that of \(P_{2}\) is \(1 / 3\). The winner will be the one who is ahead by 2 games as compared to the other player and wins at least 6 games. Now, if the player \(P_{2}\) wins first four matches, find the probability of \(P_{1}\) wining the championship.

A bag contains some white and some black balls, all combinations of balls being equally likely. The total number of balls in the bag is 10 . If three balls are drawn at random without replacement and all of them are found to be black, the probability that the bag contains 1 white and 9 black balls is a. \(14 / 55\) b. \(12 / 55\) c. \(2 / 11\) d. \(8 / 55\)

If three squares are selected at random from chessboard, then the probability that they form the letter ' \(\mathrm{L}\) ' is a. \(196^{k+4} \mathrm{C}_{3}\) b. \(49 \mathrm{~B}^{4} \mathrm{C}_{3}\) c. \(36^{\prime *} \mathrm{C}_{3}\) d. \(98^{\circ} \mathrm{C}_{3}\)

A purse contains 2 six-sided dice. One is a normal fair die, while the other has two \(1^{+} s\), two \(3^{\circ} \mathrm{s}\), and two \(5^{2}\) s. \(\mathrm{A}\) die is picked up and rolled. Because of some secret magnetic attraction of the unfair dic, there is \(75 \%\) chance of picking the unfair die and a \(25 \%\) chance of picking a fair die. The die is rolled and shows up the face 3. The probability that a fair die was picked up is a. \(1 / 7\) b. \(1 / 4\) C. \(1 / 6\) d. \(1 / 24\)

There are 20 cards. Ten of these cards have the letter ' \(T\) printed on them and the other 10 have the letter " \(T^{*}\) printed on them. If three cards are picked up at random and kept in the same order, the probability of making word IIT is a. \(4 / 27\) b. \(5 / 38\) c. \(1 / 8\) d. \(9 / 80\)

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