Conditional Probability with Three Sets

Question

I'm stuck on finding values related to C.

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Assume $$ \begin{aligned} P(A) &= 0.3\\ P(B|A) &= 0.75\\ P(B|A′) &= 0.20 \end{aligned} $$ and $$ \begin{aligned} P(C|A \cap B) &= 0.20 \\ P(C|A′ \cap B) &=0.15 \\ P(C|A \cap B′ ) &= 0.80 \\ P(C|A′ \cap B′) &= 0.90 \end{aligned} $$ The relevant parts of the question are

(a) Find $P(A \cap B \cap C)$ and

(b) Find $P(B' \cap C)$

Hint: Draw a Venn diagram.
Recall that $P(A \cap B) = P(A) \cdot P(B|A)$ and fill in the Venn diagram with the probabilities given in the problem.

This is from a homework assigment for a community college course titled, Introduction to Probability and Statistics, suitable for engineers.

The professor is using the ', the apostrophe symbol, for negation or set complement.

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I have worked out (a) and most of (b).

For (a) I got $P(A \cap B \cap C)=0.045$
I also got $P(A\cap B) =0.225$, $P(A\cap C) =0.105$,$P(B\cap C) =0.066$
I have numbers for each possible area of the Venn diagram, except for I can't find $$P(C), P( (A \cup B \cup C)') \text{ or } P(C - (A \cup B))$$ I've got multiple equations written using two versions of Bayes theorem and the definition of conditional probability for various A, B, & C. Of course to solve for two variables, I need two independent equations. Likewise, three variables needs three independent equations, etc. However, my equations all seem dependent on each other. Trying to solve a system of dependent equations leads in circles or to infinitely many solutions.

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Some context and motivation- I have a B.S. in Math. This question is actually from a student I'm tutoring. I promised him an answer, so am hoping for some help. The student is not aware of the Wronskian or tests for independence of equations, so a simpler method would be better for him if possible.

Answer 1

Part A

The first part is quite straightforward. Simply use the provided hint twice:

$$ \begin{aligned} P\left(A\cap B\cap C\right) &=P(C\mid A\cap B)\cdot P(A\cap B)\\ &=P(C\mid A\cap B)\cdot P(B\mid A) \cdot P(A)\\ &=0.20\cdot0.75\cdot0.30\\ &=0.045 \end{aligned} $$

Part B

I am going to divide the second part into multiple steps. The first step is using the complement to obtain the probability of $A'$:

$$ \begin{aligned} P(A') &= 1-P(A)\\ &= 0.70 \end{aligned} $$

Next, we can also use complement to find the following conditional probabilities' complements:

$$ \begin{aligned} P(B'\mid A\phantom{'})&=1-P(B\mid A\phantom{'})\\ &=0.25\\\\ P(B'\mid A')&=1-P(B\mid A')\\ &=0.80\\ \end{aligned} $$

Using Bayesian formula just like the first part, we get:

$$ \begin{aligned} P(A\phantom{'}\cap B')&=P(B'\mid A\phantom{'})\cdot P(A\phantom{'})\\ &=0.25\cdot0.30\\ &=0.075\\\\ P(A'\cap B')&=P(B'\mid A')\cdot P(A')\\ &=0.80\cdot0.70\\ &=0.56 \end{aligned} $$

Last, we use $A\cup A'=S$:

$$ \begin{aligned} P(C\cap B') &=P(C\cap A\cap B') + P(C\cap A'\cap B')\\ &=P(C\mid A\cap B')\cdot P(A\cap B') + P(C\mid A'\cap B')\cdot P(A'\cap B')\\ &=0.80\cdot0.075+0.90\cdot0.56\\ &=0.564 \end{aligned} $$