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Sagot :

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].

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