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Chapter 3: Problem 12

A random experiment gave rise to the two-way contingency table shown. Use it to compute the probabilities indicated. $$ \begin{array}{|l|l|l|} \hline & R & S \\ \hline A & 0.13 & 0.07 \\ \hline B & 0.61 & 0.19 \\ \hline \end{array} $$ a. \(P(A), P(R), P(A \cap R)\) b. Based on the answer to (a), determine whether or not the events \(A\) and \(R\) are independent. c. Based on the answer to (b), determine whether or not \(P(A \mid R)\) can be predicted without any computation. If so, make the prediction. In any case, compute \(P(A \mid R)\) using the Rule for Conditional Probability.

Short Answer

Expert verified
Events A and R are not independent; \(P(A \mid R) \approx 0.176\).

Step by step solution

01

Calculate P(A)

To find \(P(A)\), sum the probabilities of all events in row \(A\). Therefore, \[P(A) = P(A \cap R) + P(A \cap S) = 0.13 + 0.07 = 0.20\]
02

Calculate P(R)

To find \(P(R)\), sum the probabilities of all events in column \(R\). Therefore, \[P(R) = P(A \cap R) + P(B \cap R) = 0.13 + 0.61 = 0.74\]
03

Identify P(A 鈭� R)

From the contingency table, the intersection of events \(A\) and \(R\) is given directly as \(P(A \cap R) = 0.13\).
04

Check for Independence

Two events \(A\) and \(R\) are independent if \(P(A \cap R) = P(A) \cdot P(R)\). First, calculate \(P(A) \times P(R) = 0.20 \times 0.74 = 0.148\). Since \(P(A \cap R) = 0.13\), which is not equal to \(0.148\), the events \(A\) and \(R\) are not independent.
05

Predict P(A | R) Without Computation

Since the events \(A\) and \(R\) are not independent, \(P(A \mid R)\) cannot be directly predicted without computation.
06

Compute P(A | R) Using Conditional Probability

Conditional probability is computed using the formula \(P(A \mid R) = \frac{P(A \cap R)}{P(R)}\). Here, \[P(A \mid R) = \frac{0.13}{0.74} \approx 0.176\]

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Key Concepts

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

Probability
Probability is a measure of how likely it is for a particular event to occur. Understanding probability is key for interpreting contingency tables and solving problems based on these data displays.
To calculate the probability of a single event happening, you would look at the total possible outcomes and the number of favorable outcomes. For example, in an experiment represented by a contingency table, each value is a probability itself.
  • To find the probability of event \(A\) occurring, sum the probabilities of all events in row \(A\). This gives us \(P(A) = P(A \cap R) + P(A \cap S) = 0.13 + 0.07 = 0.20\).
  • Similarly, to find the probability of event \(R\), sum all probabilities in column \(R\), giving \(P(R) = P(A \cap R) + P(B \cap R) = 0.13 + 0.61 = 0.74\).
Probabilities are always values between 0 and 1, where 0 indicates impossibility and 1 means certainty. The ability to calculate these basic probabilities is foundational to more complex concepts such as independence and conditional probability.
Independence of Events
Two events are considered independent if the occurrence of one event does not affect the probability of the other event.
Mathematically, events \(A\) and \(R\) are independent if and only if:\[P(A \cap R) = P(A) \cdot P(R)\]Following this logic, we can determine independence using the contingency table:
  • Calculate \(P(A \cap R)\), which from the table is 0.13.
  • Next, compute \(P(A) \times P(R)\), which equals \(0.20 \times 0.74 = 0.148\).
  • Since \(P(A \cap R)\) is not equal to \(P(A) \cdot P(R)\), we conclude that events \(A\) and \(R\) are not independent.
Understanding independence is crucial for determining how events relate to each other and for simplifying probability calculations. Lack of independence implies that knowing something about event \(R\) gives some information about event \(A\), and vice versa.
Conditional Probability
Conditional probability explores the likelihood of an event occurring given that another event has already occurred.
This is crucial when working with dependent events, as it adjusts the probability based on the given condition.The formula to compute the conditional probability of \(A\) given \(R\) is:\[P(A \mid R) = \frac{P(A \cap R)}{P(R)}\]Let's break this down with our contingency table:
  • The probability of both \(A\) and \(R\) occurring is \(P(A \cap R) = 0.13\).
  • The probability of \(R\) alone is \(P(R) = 0.74\).
  • Therefore, \(P(A \mid R) = \frac{0.13}{0.74} \approx 0.176 \).
This means that given event \(R\) has occurred, there's an approximately 17.6% chance that event \(A\) also occurs. Understanding and applying conditional probability is fundamental in scenarios where events influence each other's likelihood.

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Most popular questions from this chapter

A box contains 16 white and 16 black marbles. Construct a sample space for the experiment of randomly drawing out, with replacement, three marbles in succession and noting the color each time. (To draw "with replacement" means that each marble is put back before the next marble is drawn.)

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A tourist who speaks English and German but no other language visits a region of Slovenia. If \(35 \%\) of the residents speak English, \(15 \%\) speak German, and \(3 \%\) speak both English and German, what is the probability that the tourist will be able to talk with a randomly encountered resident of the region?

A manufacturer examines its records over the last year on a component part received from outside suppliers. The breakdown on source (supplier \(A,\) supplier \(B\) ) and quality \((H:\) high \(, U:\) usable, \(D:\) defective \()\) is shown in the two-way contingency table. $$ \begin{array}{|c|c|c|c|} \hline & H & U & D \\ \hline A & 0.6937 & 0.0049 & 0.0014 \\ \hline B & 0.2982 & 0.0009 & 0.0009 \\ \hline \end{array} $$ The record of a part is selected at random. Find the probability of each of the following events. a. The part was defective. b. The part was either of high quality or was at least usable, in two ways: (i) by adding numbers in the table, and (ii) using the answer to (a) and the Probability Rule for Complements. c. The part was defective and came from supplier \(B\). d. The part was defective or came from supplier \(B\), in two ways: by finding the cells in the table that correspond to this event and adding their probabilities, and (ii) using the Additive Rule of Probability.
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