Answer:
Hence,
P(A∪B)=0.89
Step-by-step explanation:
As we are given that:
A and B are dependent events.
If P(A|B)=0.55 and P(B)=0.2 .
AS we know that:
P(A)+P(B)=1
⇒ P(A)+0.2=1
⇒ P(A)=1-0.2
⇒ P(A)=0.8.
Also we know that:
[tex]P(A|B)=\dfrac{P(A\bigcap B)}{P(B)}\\\\\\i.e.\\\\0.55=\dfrac{P(A\bigcap B)}{0.2}\\\\P(A\bigcap B)=0.55\times 0.2\\\\P(A\bigcap B)=0.11[/tex]
Hence,
[tex]P(A\bigcup B)=P(A)+P(B)-P(A\bigcap B)\\\\P(A\bigcup B)=0.8+0.2-0.11\\\\P(A\bigcup B)=0.89[/tex]
Hence,
P(A∪B)=0.89
Answer:
[tex]P(A \cup B)[/tex] = 0.89
Step-by-step explanation:
Using the formula:
[tex]P(\frac{A}{B}) = \frac{P(A \cap B)}{P(B)}[/tex] ....[1]
where, A and B are events.
As per the statement:
Suppose A and B are dependent events.
If [tex]P(\frac{A}{B})=0.55[/tex] and P(B) = 0.2
Substitute these in [1] we get;
[tex]0.55 = \frac{P(A \cap B)}{0.2}[/tex]
Multiply both sides by 0.2 we have;
[tex]0.11 = P(A \cap B)[/tex]
We know that:
[tex]P(A)+P(B) = 1[/tex]
⇒[tex]P(A) = 1-P(B)[/tex]
⇒[tex]P(A) = 1-0.2 = 0.8[/tex]
We have to find [tex]P(A \cup B)[/tex]
[tex]P(A \cup B) = P(A)+P(B) -P(A \cap B)[/tex]
Substitute the given values we have;
[tex]P(A \cup B) = 0.8+0.2-0.11 = 1.0-0.11 = 0.89[/tex]
Therefore, the value of [tex]P(A \cup B)[/tex] is, 0.89
