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Used the given information to determine the probability below round solution with 3 decimal places

Sagot :

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]