Question

section i: multiple choice select the best answer for each question. questions t5.1 to t5.3 refer to the following setting. a group of 125 truck owners were asked about the brand of truck they owned and whether or not the truck has four - wheel drive. the results are summarized in the two - way table below. suppose we randomly select one of these truck owners. t5.1 what is the probability that the person owns a dodge or has four - wheel drive? (a) 20/80 (b) 20/125 (c) 80/125 (d) 90/125 (e) 110/125 t5.2 what is the probability that the person owns a chevy, given that the truck has four - wheel drive? (a) 32/50 (b) 32/80 (c) 32/125 (d) 50/125 (e) 80/125 t5.3 which one of the following is true about the events \owners truck is a chevy\ and \owners truck has four - wheel drive\? (a) these two events are mutually exclusive and independent. (b) these two events are mutually exclusive, but not independent. (c) these two events are not mutually exclusive, but they are independent. (d) these two events are neither mutually exclusive nor independent. (e) these two events are mutually exclusive, but we do not have enough information to determine if they are independent.

Explanation:

T5.1

Step1: Calculate the number of people who own a Dodge or have four - wheel drive

Use the formula \(n(A\cup B)=n(A)+n(B)-n(A\cap B)\).
Let \(A\) be the event of owning a Dodge and \(B\) be the event of having four - wheel drive.
\(n(A)=(20 + 10)=30\), \(n(B)=(28+32 + 20)=80\), \(n(A\cap B)=20\)
\(n(A\cup B)=30 + 80-20=90\)

Step2: Calculate the probability

The total number of truck owners \(N = 125\)
The probability \(P=\frac{n(A\cup B)}{N}=\frac{90}{125}\)

T5.2

Step1: Calculate the number of people who have four - wheel drive

\(n(\text{four - wheel drive})=(28+32 + 20)=80\)

Step2: Calculate the number of people who own a Chevy and have four - wheel drive

\(n(\text{Chevy and four - wheel drive}) = 32\)

Step3: Calculate the conditional probability

Using the formula \(P(A|B)=\frac{P(A\cap B)}{P(B)}\), here \(A\) is owning a Chevy and \(B\) is having four - wheel drive.
\(P=\frac{32}{80}\)

T5.3

Step1: Check for mutual exclusivity

Two events \(A\) (owning a Chevy) and \(B\) (having four - wheel drive) are mutually exclusive if \(A\cap B=\varnothing\). But \(n(A\cap B) = 32
eq0\), so they are not mutually exclusive.

Step2: Check for independence

\(P(\text{Chevy})=\frac{32 + 18}{125}=\frac{50}{125}\), \(P(\text{four - wheel drive})=\frac{80}{125}\), \(P(\text{Chevy and four - wheel drive})=\frac{32}{125}\)
\(P(\text{Chevy})\times P(\text{four - wheel drive})=\frac{50}{125}\times\frac{80}{125}=\frac{4000}{15625}
eq\frac{32}{125}=\frac{4000}{15625}\) (after cross - multiplying \(32\times125 = 4000\) and \(125\times125=15625\))
So they are not independent

Answer:

T5.1: (d) \(90/125\)
T5.2: (b) \(32/80\)
T5.3: (d) These two events are neither mutually exclusive nor independent.