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Suppose I have four events and I'll call them $\ A,B,C,D $ .

This is the information I'm given:

$$\ D \subseteq C \subseteq B \ \ ,P(A) = 0.59 , \ P(A \cap B \cap C \cap D) = 0.10 \\ P(A | D) = 0.50, \ P(A \cup B \cup C \cup D) =0.8\\ P(A \cap B^c \cap C^c \cap D^c) = 0.2 \, \ \ P(A^c \cap B \cap C^c \cap D^c) = 0.1$$

$\ 1/3 $ of those that are part of three events ( $\ A, B,C,D $ ) do not belong to $\ D $ .

I need to find $\ P(A), P(B), P(C), P(D) $ . so far I got that because $\ D \subseteq C \subseteq B $ then $\ P(A \cap B \cap C \cap D) = P(A \cap D) = 0.10 \\ P(A | D) = \frac{P(A\cap D)}{P(D)} = 0.10 \Rightarrow P(D) = 0.2$

But couldn't find a way to find $\ P(B), P(C) $

bm1125
  • 1,427

1 Answers1

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Partial solution:

Since $B^c\subseteq C^c\subseteq D^c$ we have:

$$P(A \cap B^c \cap C^c \cap D^c) = 0.2\implies P(A\cap B^c)=0.2$$ so $$\boxed{P(A)-P(A\cap B)=0.2}$$ and since $B\cup C\cup D=B$ we have $$P(A\cup B) =0.8$$

so $$\boxed{P(A)+P(B)-P(A\cap B)=0.8}$$

so we get $P(B)=0.6$

From here $$ P(A^c \cap B \cap C^c) = 0.1$$ we should somehow find $P(C)$...

nonuser
  • 90,026