Chapter 15, Problem 25E

(a)

To determine

To find: the probability that a tested chicken was not contaminated with either type of bacteria.

Answer to Problem 25E

0.17

Explanation of Solution

Given:

  $\begin{array}{l}P\left(C\right)=0.18\\ P\left(S\right)=0.15\\ P\left(C\cap S\right)=0.13\end{array}$

Formula used:

  $P\left(C\cup S\right)=P\left(C\right)+P\left(S\right)-P\left(C\cap S\right)$

Calculation:

It is mention that 81% of the chickens were contaminated with campylobacter. 15% with salmonella and 13% with both

Let C = chicken is contaminated with Campylobacter

Let S = chicken is contaminated with Salmonella

Probability that a tested chicken was not contaminated with either kind of bacteria

  $\begin{array}{c}P\left(C\cup S\right)=P\left(C\right)+P\left(S\right)-P\left(C\cap S\right)\\ =100-0.83\\ =0.17\end{array}$

There is 17% possibility that chicken is not contaminated with either kind of bacteria

(b)

To determine

To Explain: that contamination with the two types of bacteria disjoint yes or not.

Answer to Problem 25E

Not disjoint

Explanation of Solution

Because $P\left(C\cap S\right)\ne 0$ therefore the contamination with the two types of are not disjoint

(c)

To determine

To Explain: that contamination with the two types of bacteria independent.

Answer to Problem 25E

Not independent

Explanation of Solution

No, they are independent

  $\begin{array}{c}P\left(C\backslash S\right)=\frac{P\left(C\cap S\right)}{P\left(S\right)}\\ =\frac{0.13}{0.15}\\ =0.867\\ P\left(C\right)=0.81\\ P\left(C\backslash S\right)\ne P\left(C\right)\end{array}$

Therefore they are not independent