Respuesta :
The conditional probability formula is
[tex]P(B|A)=\frac{P(B\cap A)}{P(A)}[/tex]which gives
[tex]P(B\cap A)=P(A)\times P(B|A)[/tex]Then, for the first question, we get
[tex]\begin{gathered} P(A\cap B)=P(B\cap A)=P(A)\times P(B|A) \\ P(A\cap B)=0.46\times0.05 \end{gathered}[/tex]which gives
[tex]P(A\cap B)=0.023[/tex]Now, for the second question, we know that, for independent events
[tex]P(A\cap B)=P(A)\times P(B)[/tex]then, we have
[tex]P(A\cap B)=0.46\times0.28[/tex]which gives
[tex]P(A\cap B)=0.129[/tex]Now, for question 3, we know that, when the events are dependent and mutually non-exclusive
[tex]\begin{gathered} P(A\cup B)=P(A)+P(B)-P(A\cap B) \\ P(A\cup B)=P(A)+P(B)-P(A)\times P(B|A) \end{gathered}[/tex]By substituting the given values, we have
[tex]\begin{gathered} P(A\cup B)=0.46+0.28-(0.46\times0.05) \\ P(A\cup B)=0.74+0.023 \\ P(A\cup B)=0.763 \end{gathered}[/tex]Finally, for the 4th question, we have
[tex]P(A\cup B)=P(A)+P(B)[/tex]which gives
[tex]\begin{gathered} P(A\cap B)=0.46+0.28 \\ P(A\cap B)=0.74 \end{gathered}[/tex]In summary, the solutions are:
Question 1:
[tex]P(A\cap B)=0.46\times0.05=0.023[/tex]Question 2:
[tex]P(A\cap B)=0.46\times0.28=0.129[/tex]Question 3:
[tex]P(A\cup B)=0.46+0.28-(0.46\times0.05)=0.763[/tex]Question 4:
[tex]P(A\cap B)=0.46+0.28=0.74[/tex]