Question

Early in 2007 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 \(81 \%\) of the chickens were contaminated with campylobacter, \(15 \%\) with salmonella, and \(13 \%\) 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

a) 0.17 b) No, they are not disjoint. c) No, they are not independent.

Step-by-step solution

Identifying the Given Data

We are given that \(P(C) = 0.81\), \(P(S) = 0.15\), and \(P(C \cap S) = 0.13\), where \(C\) represents contamination with campylobacter, and \(S\) represents contamination with salmonella.

a) Calculating the Probability of No Contamination

We want the probability that a chicken is not contaminated with either bacteria. The probability a chicken is contaminated with either bacteria or both is \(P(C \cup S)\), which is calculated using the formula \(P(C \cup S) = P(C) + P(S) - P(C \cap S) = 0.81 + 0.15 - 0.13 = 0.83\). Thus, the probability a chicken is not contaminated is \(1 - P(C \cup S) = 1 - 0.83 = 0.17\).

b) Checking if the Events are Disjoint

Two events are disjoint if they cannot happen simultaneously. Here, \(P(C \cap S) = 0.13\), which means some chickens are contaminated with both bacteria. Therefore, contamination with campylobacter and salmonella is not disjoint.

c) Checking for Independence

Two events are independent if \(P(C \cap S) = P(C) \cdot P(S)\). Calculate \(P(C) \cdot P(S) = 0.81 \cdot 0.15 = 0.1215\). Since \(P(C \cap S) = 0.13\) and \(0.13 eq 0.1215\), the events are not independent.

Key concepts

Disjoint Events

Disjoint events, also known as mutually exclusive events, are events that cannot occur at the same time. In simpler terms, if one event happens, the other cannot. For example, when you roll a die, the outcomes 1 and 2 are disjoint because both cannot occur on a single roll.

In the context of our chicken contamination problem, we are looking at the events of contamination with either campylobacter or salmonella bacteria. To determine if these are disjoint, we check if both can happen simultaneously. If they cannot happen together, they are disjoint.

In our exercise, it is given that 13% of chickens have both bacteria, meaning contamination with both is possible. Because some chickens are contaminated with both types of bacteria, these events are not disjoint. Instead, they can indeed occur simultaneously in some chickens, which is why they are not mutually exclusive.

Independent Events

Independent events are those whose occurrence does not impact each other. In other words, the occurrence of one event doesn’t change the probability of the other. For instance, flipping a coin results in heads or tails, which are independent of each other.

To check if the events of contamination with campylobacter and salmonella are independent, we use a specific condition. For events to be independent, their joint probability must equal the product of their individual probabilities. Mathematically, this is expressed as:

• \( P(C \cap S) = P(C) \cdot P(S) \)

For our case:

• \(P(C \cap S) = 0.13\)
• \(P(C) = 0.81\)
• \(P(S) = 0.15\)

Calculating their product, we get \(0.81 \times 0.15 = 0.1215\).

Since \(0.13 eq 0.1215\), the events are not independent. Thus, contamination with one bacteria does seem to affect the probability of contamination with the other.

Contamination

Contamination here refers to the presence of bacteria in food products, specifically campylobacter and salmonella in chickens. Understanding the probabilities involving contamination can help in assessing food safety and the risk of infection from the consumed chicken.

In terms of probability, assessing the chances of finding either bacteria, or both, in chickens is essential. We calculated the probability of finding contamination using the formula for unions of probabilities \( P(C \cup S) = P(C) + P(S) - P(C \cap S) \). The subtraction of the intersection \( P(C \cap S) \) is necessary because this probability is included in both \(P(C)\) and \(P(S)\).

The outcome tells us that roughly 17% of the chickens tested had neither type of bacteria, while 83% were contaminated with one or both. Knowing these probabilities helps consumers and health organizations understand the extent of risk and to take measures accordingly to minimize contamination. Keeping food safety standards high is crucial to reducing these percentages.