Question

a spinner is divided into sections that are both colored and numbered. the event that the spinner lands on a green - section is independent from the event that it lands on an odd - numbered section. which of the following sets of probabilities could be associated with this spinner? p(green)=0.25 p(odd)=0.2 p(green and odd)=0.45 p(green)=0.12 p(odd)=0.2 p(green and odd)=0.06 p(green)=0.18 p(odd)=0.2 p(green and odd)=0.036

Explanation:

Step1: Recall the formula for independent events

For two independent events \(A\) and \(B\), \(P(A\cap B)=P(A)\times P(B)\). Let \(A\) be the event of landing on a green - section and \(B\) be the event of landing on an odd - numbered section.

Step2: Check each option

Option 1

If \(P(\text{green}) = 0.18\), \(P(\text{odd})=0.2\), then \(P(\text{green}\cap\text{odd})=P(\text{green})\times P(\text{odd})=0.18\times0.2 = 0.036\)

Option 2

If \(P(\text{green}) = 0.12\), \(P(\text{odd}) = 0.2\), then \(P(\text{green}\cap\text{odd})=P(\text{green})\times P(\text{odd})=0.12\times0.2=0.024
eq0.06\)

Option 3

If \(P(\text{green}) = 0.25\), \(P(\text{odd}) = 0.2\), then \(P(\text{green}\cap\text{odd})=P(\text{green})\times P(\text{odd})=0.25\times0.2 = 0.05
eq0.45\)

Answer:

The first set of probabilities \(P(\text{green}) = 0.18\), \(P(\text{odd}) = 0.2\), \(P(\text{green and odd}) = 0.036\) could be associated with the spinner since for independent events \(P(\text{green}\cap\text{odd})=P(\text{green})\times P(\text{odd})\) holds for this set.