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Determine whether the events \(A\) and \(B\) are independent. $$ P(A)=.5, P(B)=.7, P(A \cup B)=.85 $$

Short Answer

Expert verified
We are given \(P(A)=0.5, P(B)=0.7,\) and \(P(A \cup B)=0.85\). Using the formula for the union of two events, \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\), we find \(P(A \cap B) = 0.35\). Since \(P(A \cap B) \neq P(A) \cdot P(B)\), events A and B are not independent.

Step by step solution

01

Write the formula for the probability of the union of two events

The formula for the probability of the union of two events A and B is given by: \[P(A \cup B) = P(A) + P(B) - P(A \cap B).\]
02

Plug in the given values into the formula

We know that \(P(A \cup B) = 0.85, P(A) = 0.5,\) and \(P(B) = 0.7\). We plug these values into the formula: \[0.85 = 0.5 + 0.7 - P(A \cap B).\]
03

Solve for the probability of the intersection of A and B

To find the probability of the intersection, we solve for \(P(A \cap B)\) in the equation we derived in Step 2: \[P(A \cap B) = 0.5 + 0.7 - 0.85 = 0.35.\]
04

Check if the events are independent

Now, we check if \(P(A \cap B) = P(A) \cdot P(B)\): \[P(A \cap B) = 0.35 \neq 0.5 \cdot 0.7 = 0.35.\] Since the equality does not hold, we can conclude that events A and B are not independent.

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Key Concepts

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

Probability of the Union of Two Events
Understanding the probability of the union of two events is essential in determining the likelihood that at least one of two events will happen. The union of events A and B, denoted as \( A \bigcup B \), encompasses all outcomes that belong to either A, B, or both. The probability of this union is expressed as \( P(A \bigcup B) = P(A) + P(B) - P(A \bigcap B) \).

This formula essentially adds up the probabilities of each event occurring and then subtracts the probability of them happening simultaneously to account for overlapping outcomes, which would otherwise be double-counted. It's similar to how you would calculate the crowd at a concert that has fans of two bands performing. You count fans of band A, fans of band B, and then subtract the fans who like both to avoid counting them twice.

In our exercise, given \( P(A) = 0.5 \) and \( P(B) = 0.7 \), the assumed probability for the union was \( P(A \bigcup B) = 0.85 \). By applying these values to the formula, we can explore the relationship between A and B further.
Probability of the Intersection of Events
The probability of the intersection of two events, \( A \bigcap B \), is fundamental when you want to find the probability that both events occur simultaneously. Imagine you're trying to find out how likely it is to roll a die and get an even number (event A) while also flipping a coin and getting heads (event B). The intersection is the occurrence of both an even die roll and the coin landing on heads.

To compute this, we already have the relationship \( P(A \bigcap B) = P(A) + P(B) - P(A \bigcup B) \). Using the values from our earlier equation, we deduced that \( P(A \bigcap B) = 0.35 \). This step is much like solving a puzzle by fitting the pieces of individual event probabilities together to see what they collectively reveal about the potential overlap.
Conditional Probability
When talking about conditional probability, you're dealing with the probability of an event A, given that another event B has already occurred. It is denoted by \( P(A | B) \), and it answers questions like, 'What are the chances of drawing an ace from a deck of cards given that we know the card is black?'

The conditional probability can be understood through the formula \( P(A | B) = \frac{P(A \bigcap B)}{P(B)} \) if B has a probability greater than zero. This equation essentially states that the probability of A happening, given B, is the probability of both A and B happening divided by the probability of B on its own.

In the context of independent events, if knowing that B occurs does not change the probability of A, then A and B are independent, and \( P(A | B) = P(A) \). However, in the exercise we saw that \( P(A \bigcap B) = 0.35 \) does not equal \( P(A) \times P(B) = 0.5 \times 0.7 \), indicating that A and B are not independent events. Their conditional probabilities would indeed differ, illustrating how these concepts interlink to portray a deeper understanding of event relationships.

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Most popular questions from this chapter

Applicants for temporary office work at Carter Temporary Help Agency who have successfully completed a typing test are then placed in suitable positions by Nancy Dwyer and Darla Newberg. Employers who hire temporary help through the agency return a card indicating satisfaction or dissatisfaction with the work performance of those hired. From past experience it is known that \(80 \%\) of the employees placed by Nancy are rated as satisfactory, and \(70 \%\) of those placed by Darla are rated as satisfactory. Darla places \(55 \%\) of the temporary office help at the agency and Nancy the remaining \(45 \%\). If a Carter office worker is rated unsatisfactory, what is the probability that he or she was placed by Darla?

The accompanying data were obtained from the financial aid office of a certain university: \begin{tabular}{lccc} \hline & \multicolumn{3}{c} { Not } & \\ & Receiving Financial Aid & Receiving Financial Aid & Total \\ \hline Undergraduates & \(4.222\) & 3,898 & 8,120 \\ \hline Graduates & \(1.879\) & 731 & 2,610 \\ \hline Total & \(6.101\) & 4,629 & 10,730 \\ \hline \end{tabular} Let \(A\) be the event that a student selected at random from this university is an undergraduate student, and let \(B\) be the event that a student selected at random is receiving financial aid. a. Find each of the following probabilities: \(P(A), P(B)\), \(P(A \cap B), P(B \mid A)\), and \(P\left(B \mid A^{\circ}\right)\) b. Are the events \(A\) and \(B\) independent events?

A study was conducted among a certain group of union members whose health insurance policies required second opinions prior to surgery. Of those members whose doctors advised them to have surgery, \(20 \%\) were informed by a second doctor that no surgery was needed. Of these, \(70 \%\) took the second doctor's opinion and did not go through with the surgery. Of the members who were advised to have surgery by both doctors, \(95 \%\) went through with the surgery. What is the probability that a union member who had surgery was advised to do so by a second doctor?

An experiment consists of randomly selecting one of three coins, tossing it, and observing the outcome-heads or tails. The first coin is a two-headed coin, the second is a biased coin such that \(P(\mathrm{H})=.75\), and the third is a fair coin. a. What is the probability that the coin that is tossed will show heads? b. If the coin selected shows heads, what is the probability that this coin is the fair coin?

The Office of Admissions and Records of a large western university released the accompanying information concerning the contemplated majors of its freshman class:3 $$\begin{array}{lccc} \text { Major } & \text {This Major, \% } & \text { Females, \% } & \text { Males, \% } \\ \hline \text { Business } & 24 & 38 & 62 \\ \hline \text { Humanities } & 8 & 60 & 40 \\ \hline \text { Education } & 8 & 66 & 34 \\ \hline \text { Social science } & 7 & 58 & 42 \\ \hline \text { Natural sciences } & 9 & 52 & 48 \\ \hline \text { Other } & 44 & 48 & 52 \\ \hline \end{array}$$ What is the probability that a. A student selected at random from the freshman class is a female? b. A business student selected at random from the fresh- man class is a male? c. A female student selected at random from the freshman class is majoring in business?

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