We are given the following:
[tex]\begin{gathered} P\mleft(A\mright)=0.6 \\ P\mleft(A,orB\mright)=0.98 \\ P(A,\text{and }B)=0.02 \\ \\ \text{For Mutually }Exclusive,\text{ we have:} \\ P\mleft(A,orB\mright)=P\mleft(A\mright)+P\mleft(B\mright)=0.98 \\ \Rightarrow0.6+P(B)=0.98 \\ P\mleft(B\mright)=0.98-0.6=0.38 \\ P\mleft(B\mright)=0.38 \\ \\ \text{For Mutually }Inclusive,\text{ we have:} \\ P(A,orB)=P(A)+P(B)-P(A,\text{and }B) \\ 0.98=0.6+P(B)-0.02 \\ P(B)=0.98+0.02-0.6 \\ P(B)=1-0.6 \\ P(B)=0.4 \\ \\ For\text{ Independent Events, we have:} \\ P(A,\text{and }B)=P(A)\times P(B) \\ 0.02=0.6\times P(B) \\ P(B)=\frac{0.02}{0.6} \\ P(B)=0.033 \end{gathered}[/tex]The answer for P(B) would therefore be determined by the type of events that A & B are
