Respuesta :
Answer:
a) P(A or B) = 0.8
b) P(A or B) = 0.7
Step-by-step explanation:
We have two events:
Event A
Event B
We have that:
[tex]P(A) = P(a) + P(A \cap B)[/tex]
In which P(a) is the probability that only a happens and [tex]A \cap B[/tex] is the probability that both A and B happen.
By the same logic, we have that:
[tex]P(B) = P(b) + P(A \cap B)[/tex]
P(A or B) is
[tex]P(A \cup B) = P(a) + P(b) + P(A \cap B)[/tex]
We have that:
[tex]P(A) = 0.3, P(B) = 0.5[/tex]
(a) If A and B are mutually exclusive events, compute P(A or B).
Since A and B are mutually exclusive events, we have that:
[tex]P(A \cap B) = 0[/tex]
So
[tex]P(A) = 0.3[/tex]
[tex]P(A) = P(a) + P(A \cap B)[/tex]
[tex]0.3 = P(a) + 0[/tex]
[tex]P(a) = 0.3[/tex]
----------------
[tex]P(B) = 0.5[/tex]
[tex]P(B) = P(b) + P(A \cap B)[/tex]
[tex]0.5 = P(b) + 0[/tex]
[tex]P(b) = 0.5[/tex]
Or operation:
[tex]P(A \cup B) = P(a) + P(b) + P(A \cap B) = 0.3 + 0.5 + 0 = 0.8[/tex]
(b) If P(A and B) = 0.1, compute P(A or B).
[tex]P(A \cap B) = 0.1[/tex]
So
[tex]P(A) = 0.3[/tex]
[tex]P(A) = P(a) + P(A \cap B)[/tex]
[tex]0.3 = P(a) + 0.1[/tex]
[tex]P(a) = 0.2[/tex]
----------------
[tex]P(B) = 0.5[/tex]
[tex]P(B) = P(b) + P(A \cap B)[/tex]
[tex]0.5 = P(b) + 0.1[/tex]
[tex]P(b) = 0.4[/tex]
Or operation:
[tex]P(A \cup B) = P(a) + P(b) + P(A \cap B) = 0.2 + 0.4 + 0.1 = 0.7[/tex]
