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

User Brooksbp
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1 Answer

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The conditional probability formula is


P(B|A)=(P(B\cap A))/(P(A))

which gives


P(B\cap A)=P(A)* P(B|A)

Then, for the first question, we get


\begin{gathered} P(A\cap B)=P(B\cap A)=P(A)* P(B|A) \\ P(A\cap B)=0.46*0.05 \end{gathered}

which gives


P(A\cap B)=0.023

Now, for the second question, we know that, for independent events


P(A\cap B)=P(A)* P(B)

then, we have


P(A\cap B)=0.46*0.28

which gives


P(A\cap B)=0.129

Now, for question 3, we know that, when the events are dependent and mutually non-exclusive


\begin{gathered} P(A\cup B)=P(A)+P(B)-P(A\cap B) \\ P(A\cup B)=P(A)+P(B)-P(A)* P(B|A) \end{gathered}

By substituting the given values, we have


\begin{gathered} P(A\cup B)=0.46+0.28-(0.46*0.05) \\ P(A\cup B)=0.74+0.023 \\ P(A\cup B)=0.763 \end{gathered}

Finally, for the 4th question, we have


P(A\cup B)=P(A)+P(B)

which gives


\begin{gathered} P(A\cap B)=0.46+0.28 \\ P(A\cap B)=0.74 \end{gathered}

In summary, the solutions are:

Question 1:


P(A\cap B)=0.46*0.05=0.023

Question 2:


P(A\cap B)=0.46*0.28=0.129

Question 3:


P(A\cup B)=0.46+0.28-(0.46*0.05)=0.763

Question 4:


P(A\cap B)=0.46+0.28=0.74

User Ivan Pierre
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