Question

select the correct answer. the venn diagram shows event a and event b comprised of outcomes from the same sample space. the probability of event a is given, as well as the probability of neither event a nor event b. what is the probability of event b? a. 0.5 b. 0.4 c. 0.2 d. 0.6

Explanation:

Step1: Recall probability rule

The sum of probabilities of all possible outcomes in the sample - space is 1.

Step2: Identify given probabilities

We know that \(P(A)=0.2\) and \(P(B) = 0.2\), and we want to find the probability of neither \(A\) nor \(B\), denoted as \(P(\overline{A\cup B})\).

Step3: Use the formula \(P(A\cup B)=P(A)+P(B)\) (since \(A\) and \(B\) are mutually - exclusive as there is no overlap in the Venn - diagram)

\(P(A\cup B)=0.2 + 0.2=0.4\).

Step4: Use the complement rule \(P(\overline{A\cup B})=1 - P(A\cup B)\)

\(P(\overline{A\cup B})=1-0.4 = 0.6\). But we made a wrong start. We want \(P(B)\). Since the sum of \(P(A)\), \(P(B)\) and \(P(\overline{A\cup B})\) is 1, and we know \(P(A) = 0.2\) and \(P(\overline{A\cup B})\) (probability of neither \(A\) nor \(B\)) is what we were initially mis - calculating. If we assume the probability of the non - \(A\) and non - \(B\) part is \(x\), \(P(A)+P(B)+x = 1\). Given \(P(A) = 0.2\) and we want to find \(P(B)\). Since the Venn diagram shows \(P(B)=0.2\) and the probability of neither \(A\) nor \(B\) is \(1-(0.2 + 0.2)=0.6\). But the question asks for \(P(B)\) which from the Venn diagram is clearly \(0.2\). If we want to find the probability of neither \(A\) nor \(B\):

1. First, note that for mutually exclusive events \(A\) and \(B\) (no overlap in Venn diagram), the probability of \(A\cup B\) is \(P(A\cup B)=P(A)+P(B)\). Here \(P(A) = 0.2\) and \(P(B)=0.2\), so \(P(A\cup B)=0.2 + 0.2=0.4\).
2. Then, the probability of the complement of \(A\cup B\) (neither \(A\) nor \(B\)) is \(P(\overline{A\cup B})=1 - P(A\cup B)=1 - 0.4=0.6\). But we want \(P(B)\) which is \(0.2\).

So the probability of event \(B\) is \(0.2\).

Answer:

C. 0.4