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3 years ago

If A and B are two independent events with P(A)=2÷3 and P(B) =1÷5, find P(AUB).

Mathematics

1 answer:

iren [92.7K]3 years ago

3 0

Answer:

$for \: independent \: events : \\ P(AUB) = P(A) + P(B) \\ P(AUB) = \frac{2}{3} + \frac{1}{5} \\ = \frac{13}{15}$

Step-by-step explanation:

P(AUB) is P(A) + P(B) because P(AnB) is zero.

$P(AUB) = P(A) + P(B) - P(A{ \huge{n} }B) \\ P(AUB) = P(A) + P(B) + 0$

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3 years ago

Read 2 more answers

Find the exact value of tan (arcsin (two fifths)). For full credit, explain your reasoning.

Hitman42 [59]

$\bf sin^{-1}(some\ value)=\theta \impliedby \textit{this simply means}
\\\\\\
sin(\theta )=some\ value\qquad \textit{now, also bear in mind that}
\\\\\\
sin(\theta)=\cfrac{opposite}{hypotenuse}\qquad 
\qquad 
% tangent
tan(\theta)=\cfrac{opposite}{adjacent}\\\\
-------------------------------\\\\$

$\bf sin^{-1}\left( \frac{2}{5} \right)=\theta \impliedby \textit{this simply means that}
\\\\\\
sin(\theta )=\cfrac{2}{5}\cfrac{\leftarrow opposite}{\leftarrow hypotenuse}\qquad \textit{now let's find the adjacent side}
\\\\\\
\textit{using the pythagorean theorem}\\\\
c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a\qquad 
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite\\
\end{cases}$

$\bf \pm \sqrt{5^2-2^2}=a\implies \pm\sqrt{21}=a
\\\\\\
\textit{we don't know if it's +/-, so we'll assume is the + one}\quad \sqrt{21}=a\\\\
-------------------------------\\\\
tan(\theta)=\cfrac{opposite}{adjacent}\qquad \qquad tan(\theta)=\cfrac{2}{\sqrt{21}}
\\\\\\
\textit{and now, let's rationalize the denominator}
\\\\\\
\cfrac{2}{\sqrt{21}}\cdot \cfrac{\sqrt{21}}{\sqrt{21}}\implies \cfrac{2\sqrt{21}}{(\sqrt{21})^2}\implies \cfrac{2\sqrt{21}}{21}
$

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3 years ago

If A and B are two independent events with P(A)=2÷3 and P(B) =1÷5, find P(AUB).​

If A and B are two independent events with P(A)=2÷3 and P(B) =1÷5, find P(AUB).