Find P((A∪B)'): Mutually Excl. & P(A)=P(B) = 1/5

[mathlearn]

If the two events A & B are mutually exclusive and $P(A) = P(B) =\frac{1}{5}$, then find $P((A∪B)')$.

I cannot recall anything on this, help would be appreciated :)

 

[HallsofIvy]

$(A\cup B)'= A' \cap B'$

 

[mathlearn]

$(A\cup B)'= A' \cap B'$

$P(A') = P(B') =\frac{4}{5}$

With that known ,

$ A' \cap B' = \frac{1}{5}$

Correct? :)

 

[I like Serena]

If the two events A & B are mutually exclusive and $P(A) = P(B) =\frac{1}{5}$, then find $P((A∪B)')$.

I cannot recall anything on this, help would be appreciated :)

Mutually exclusive means $P(A\cup B)=P(A)+P(B)$.
We can combine it with the complement rule that states that $P(E')=1-P(E)$. (Thinking)

I'm afraid I don't see what we can do with $A'\cap B'$.

 

[mathlearn]

Mutually exclusive means $P(A\cup B)=P(A)+P(B)$.
We can combine it with the complement rule that states that $P(E')=1-P(E)$. (Thinking)

I'm afraid I don't see what we can do with $A'\cap B'$.

$P(A\cup B)=P(A)+P(B)$
$P(A\cup B)=P(\frac{1}{5})+P(\frac{1}{5})$
$P(A\cup B)=P(\frac{2}{5})$

$P((A∪B)')$ This implies that everything except the addition of the probability of the two events $P(\frac{3}{5})$, Correct ?

 

[I like Serena]

$P(A\cup B)=P(A)+P(B)$
$P(A\cup B)=P(\frac{1}{5})+P(\frac{1}{5})$
$P(A\cup B)=P(\frac{2}{5})$

That should be:

$P(A\cup B)=P(A)+P(B)$
$P(A\cup B)=\frac{1}{5}+\frac{1}{5}$
$P(A\cup B)=\frac{2}{5}$

since $P$ is a function of an event that yields a number. (Nerd)

$P((A∪B)')$ This implies that everything except the addition of the probability of the two events $P(\frac{3}{5})$, Correct ?

Yep. (Nod)
More specifically:

$P((A∪B)') = 1 - P(A∪B) = 1 - \frac 25 = \frac 35$

 

[mathlearn]

That should be:

$P(A\cup B)=P(A)+P(B)$
$P(A\cup B)=\frac{1}{5}+\frac{1}{5}$
$P(A\cup B)=\frac{2}{5}$

since $P$ is a function of an event that yields a number. (Nerd)
Yep. (Nod)
More specifically:

$P((A∪B)') = 1 - P(A∪B) = 1 - \frac 25 = \frac 35$

(Star) What a fantastic explanation , Thank you very much (Nerd)