Respuesta :

LammettHash
By the definition of conditional probability, you have

[tex]\mathbb P(B|A)=\dfrac{\mathbb P(B\cap A)}{\mathbb P(A)}\implies\mathbb P(A\cap B)=0.2\times0.5=0.1[/tex]

[tex]\mathbb P(A|B)=\dfrac{\mathbb P(A\cap B)}{\mathbb P(B)}\implies\mathbb P(B)=\dfrac{0.1}{0.25}=1[/tex]

[tex]\mathbb P(\neg B|\neg A)=\dfrac{\mathbb P(\neg B\cap\neg A)}{\mathbb P(\neg A)}[/tex]

Recall that

[tex]\mathbb P(\neg B\cap\neg A)=\mathbb P(\neg(B\cup A))=1-\mathbb P(B\cup A)[/tex]

and that

[tex]\mathbb P(B\cup A)=\mathbb P(B)+\mathbb P(A)-\mathbb P(B\cap A)[/tex]

This means

[tex]\mathbb P(\neg B|\neg A)=\dfrac{1-\mathbb P(B)-\mathbb P(A)+\mathbb P(B\cap A)}{1-\mathbb P(A)}=\dfrac{1-0.4-0.5+0.1}{1-0.5}=0.1[/tex]

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