Question

asked Sep 3, 2023 192k views

1 Answer

Ivan Pierre · Answer 1 · 2023-09-08T01:45:37+0000

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:

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

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