We have two events A and B.
We know that:
P(B) = 0.5
P(A|B) = 0.4
P(A∩B') = 0.4
i) We have to calculate P(A∩B).
We can relate P(A∩B) with the other probabilities knowing that:
[tex](A\cap B)\cup(A\cap B^{\prime})=A[/tex]So we can write:
[tex]P(A\cap B)+P(A\cap B^{\prime})=P(A)[/tex]We know P(A∩B') but we don't know P(A), so this approach is not useful in this case.
We can try with the conditional probability relating P(A∩B) as:
[tex]P(A|B)=\frac{P(A\cap B)}{P(B)}[/tex]In this case, we can use this to calculate P(A∩B) as:
[tex]\begin{gathered} P(A\cap B)=P(A|B)P(B) \\ P(A\cap B)=0.4*0.5 \\ P(A\cap B)=0.2 \end{gathered}[/tex]ii) We have to calculate P(A) now.
We can use the first equation we derive to calculate it:
[tex]\begin{gathered} P(A)=P(A\cap B)+P(A\cap B^{\prime}) \\ P(A)=0.2+0.4 \\ P(A)=0.6 \end{gathered}[/tex]iii) We have to calculate P(A∪B).
We can use the expression:
[tex]\begin{gathered} P(A\cup B)=P(A)+P(B)-P(A\cap B) \\ P(A\cup B)=0.6+0.4-0.2 \\ P(A\cup B)=0.8 \end{gathered}[/tex]iv. We can now calculate P(A|B') as:
[tex]\begin{gathered} P(A)=P(A|B)+P(A|B^{\prime}) \\ P(A|B^{\prime})=P(A)-P(A|B) \\ P(A|B^{\prime})=0.6-0.4 \\ P(A|B^{\prime})=0.2 \end{gathered}[/tex]b) We now have to find if A and B are independent events.
To do that we have to verify this conditions:
[tex]\begin{gathered} 1)P(A|B)=P(A) \\ 2)P(B|A)=P(B) \\ 3)P(A\cap B)=P(A)*P(B) \end{gathered}[/tex]We can check for the first condition, as we already know the value:
[tex]\begin{gathered} P(A|B)=0.4 \\ P(A)=0.6 \\ =>P(A|B)P(A) \end{gathered}[/tex]Then, the events are not independent.
Answer:
i) P(A∩B) = 0.2
ii) P(A) = 0.6
iii) P(A∪B) = 0.8
iv) P(A|B') = 0.2
b) The events are not independent.
