Given:
For two events X and Y,
[tex]P(X)=\dfrac{2}{3}[/tex]
[tex]P(Y)=\dfrac{2}{5}[/tex]
[tex]P(X|Y)=\dfrac{1}{5}[/tex]
To find:
The probabilities [tex]P(Y\cap X), P(Y)\cdot P(X)[/tex].
Solution:
Using the conditional probability:
[tex]P(X|Y)=\dfrac{P(Y\cap X)}{P(Y)}[/tex]
[tex]P(X|Y)\times P(Y)=P(Y\cap X)[/tex]
Substituting the given values, we get
[tex]\dfrac{1}{5}\times \dfrac{2}{5}=P(Y\cap X)[/tex]
[tex]\dfrac{2}{25}=P(Y\cap X)[/tex]
And,
[tex]P(Y)\times P(X)=\dfrac{2}{5}\times \dfrac{2}{3}[/tex]
[tex]P(Y)\times P(X)=\dfrac{4}{15}[/tex]
Therefore, the required probabilities are [tex]P(Y\cap X)=\dfrac{2}{25}[/tex] and [tex]P(Y)\times P(X)=\dfrac{4}{15}[/tex].
