Question

The probability that a randomly selected worker primarily drives a bicycle to work is 0.708. The probability that a randomly selected worker takes public transportation is 0.096.

Answer 1

Given:

The probability that a randomly selected worker primarily drives a bicycle to work is,

`P\mleft(A\mright)=0.708`

The probability that a randomly selected worker takes public transportation is,

`P(B)=0.096`

Step-by-step explanation:

a) To find: The probability that a randomly selected worker primarily drives a bicycle or public transportation to work

`\begin{gathered} P(A\cup B)=P(A)+P(B) \\ =0.708+0.096 \\ =0.804 \end{gathered}`

Thus, the answer is 0.804.

b) To find: The probability that a randomly selected worker primarily drives neither a bicycle nor public transportation to work

`\begin{gathered} 1-P(A\cup B)=1-0.804 \\ =0.196 \end{gathered}`

Thus, the answer is 0.196.

c) To find: The probability that a randomly selected worker does not drive a bicycle to work

`\begin{gathered} P(\bar{A})=1-P(A) \\ =1-0.708 \\ =0.292 \end{gathered}`

Thus, the answer is 0.292.

d) To check: The probability that a randomly selected worker walks to work is 0.30.

We know that the total probability is 1.

`\begin{gathered} P(A)+P(B)+P(C)=1 \\ 0.708+0.096+0.292=1 \\ 1.096=1 \\ \text{But, }1.096>1 \end{gathered}`

So, the answer is,

No, the probability a worker primarily drives a bicycle, walks, or public transportation would be greater than 1.