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Chapter 14: Problem 35

Early in \(2010,\) Consumer Reports published the results of an extensive investigation of broiler chickens purchased from food stores in 23 states. Tests for bacteria in the meat showed that \(62 \%\) of the chickens were contaminated with campylobacter, \(14 \%\) with salmonella, and \(9 \%\) with both. a) What's the probability that a tested chicken was not contaminated with either kind of bacteria? b) Are contamination with the two kinds of bacteria disjoint? Explain. c) Are contamination with the two kinds of bacteria independent? Explain.

Short Answer

Expert verified
a) 0.33; b) No, they are not disjoint; c) No, they are not independent.

Step by step solution

01

Identify the Known Probabilities

We know: \( P(C) = 0.62 \) (probability of contamination with campylobacter), \( P(S) = 0.14 \) (probability of contamination with salmonella), and \( P(C \cap S) = 0.09 \) (probability of contamination with both bacteria).
02

Use Inclusion-Exclusion Principle

To find the probability that a chicken is contaminated with either type of bacteria, use the formula for union of two events: \( P(C \cup S) = P(C) + P(S) - P(C \cap S) = 0.62 + 0.14 - 0.09 = 0.67 \).
03

Calculate Probability of No Contamination

The probability that a chicken is not contaminated with either type of bacteria is the complement of the union: \( P((C \cup S)^c) = 1 - P(C \cup S) = 1 - 0.67 = 0.33 \).
04

Check If Events Are Disjoint

Two events are disjoint if they cannot occur simultaneously. Since \( P(C \cap S) = 0.09 eq 0 \), the events are not disjoint because some chickens are contaminated with both bacteria.
05

Check If Events Are Independent

Two events are independent if \( P(C \cap S) = P(C) \times P(S) \). Calculate: \( 0.62 \times 0.14 = 0.0868 \). Since \( P(C \cap S) = 0.09 \), the events are not independent because \( 0.09 eq 0.0868 \).

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Key Concepts

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

Inclusion-Exclusion Principle
The Inclusion-Exclusion Principle is a vital tool in probability that helps find the probability of the union of two or more events. It corrects the straightforward addition of probabilities by subtracting the part where events overlap.

When calculating the probability of a chicken being contaminated with either campylobacter or salmonella, one might initially add the probabilities of each contamination: \( P(C) \) and \( P(S) \).

However, this approach would count the probability of dual contamination twice. To adjust, use the formula for two events:
  • \( P(C \cup S) = P(C) + P(S) - P(C \cap S) \)
This method ensures accurate results by properly accounting for overlapping probabilities.
Disjoint Events
Disjoint events, also known as mutually exclusive events, occur when two or more events cannot happen simultaneously.

In simpler terms, if event A happens, then event B cannot, and vice versa.

In our chicken contamination context, the question was whether being contaminated by either campylobacter or salmonella simultaneously is possible.

Since chickens have been found to be contaminated with both bacteria types (\( P(C \cap S) = 0.09 \)), the two types of contamination are not disjoint events. Disjoint events would mean that \( P(C \cap S) = 0 \).
Independent Events
Independent events in probability mean that the occurrence of one event does not affect the probability of the other.

In essence, for two events, C and S, they are independent if:
  • \( P(C \cap S) = P(C) \times P(S) \)
To determine if chicken contamination with both bacteria types is independent, compare the calculated intersection probability (\( P(C \cap S) \)) with the product of the individual probabilities (\( P(C) \times P(S) \)).

Here, \( 0.62 \times 0.14 = 0.0868 \), but the actual joint probability is \( 0.09 \). Since these values do not match, contamination events for these bacteria are not independent.
Complement Rule
The Complement Rule is a fundamental principle in probability used to determine the likelihood of an event not occurring.

This is essential when calculating probabilities for events dealing with 'at least' or 'not happening'.

To find the probability of a chicken not being contaminated with either bacteria, use:
  • \( P((C \cup S)^c) = 1 - P(C \cup S) \)
Substitute the value of \( P(C \cup S) \) from our previous calculation of contamination with either bacteria, which is \( 0.67 \), and find that the probability of no contamination is \( 0.33 \).

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Most popular questions from this chapter

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