Generating Sets From Given Information

The problem I am working on is:

The three most popular options on a certain type of new car are a built-in GPS (A), a sunroof (B), and an automatic transmission (C). If 40% of all purchasers request A, 55% request B, 70% request C, 63% request A or B, 77% request A or C, 80% request B or C, and 85% request A or B or C, determine the probabilities of the following events. [Hint: “Aor B” is the event that at least one of the two options is requested; try drawing a Venn diagram and labeling all regions.]

a.The next purchaser will request at least one of the three options.

b.The next purchaser will select none of the three options.

c. The next purchaser will request only an automatic transmission and not either of the other two options.

d.The next purchaser will select exactly one of these three options.

I am absolutely positive that $P(A)=40\%$, $P(B)=55\%$, $P(C)=70\%$, and $P(A \cup B \cup C)=85\%$. However, the pieces of data I am not quite certain about are $P(A \cup B \cap C')=63\%$, $P(A \cup C \cap B')=77\%$, $P(B \cup C \cap A')=80\%$, do these values correspond to the rest of the data? If so, then I seems nearly impossible to be able to generate the Venn Diagram. Could someone help?

EDIT: What I am having a difficult time interpreting is, when they say in the question, "...63% request A or B." To me, that says only A or only B; and under this interpretation I would write $P(A \cup B \cap C')=63\%$. Under André Nicolas' interpretation, "63% request A or B," means $P(A \cup B)=63\%$. If it is the case that André Nicolas is correct, then it seems like they should have stated in the question, "63% request A or B, A and C, B and C, or A and B and C."

Also, I solved the problem under André Nicolas' assumption, and for part d), I know the answer but I am sure how to put in it math symbols. How would I do that?

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Here is a start:

a) The answer to this is written down in the data: It is $\Pr(A\cup B\cup C)$.

b) This can be immediately written down from the answer to a).

As for the rest of the probabilities, we need to work. It can be figured out completely from a Venn diagram. But since one cannot point to a picture on a blackboard, what follows will be formula-heavy.

Recall that $$\Pr(X\cup Y)=\Pr(X)+\Pr(Y)-\Pr(X\cap Y).\tag{1}$$

We know $\Pr(A)$, $\Pr(B)$, and $\Pr(A\cup B)$. Using Formula $(1)$, we can see that $0.63=0.40+0.55-\Pr(A\cap B)$. That gives $\Pr(A\cap B)=0.32$.

Similarly, we can find $\Pr(A\cap C)$ and $\Pr(B\cap C)$ from the given information.

Recall also that

$$\Pr(A\cup B\cup C)=\Pr(A)+\Pr(B)+\Pr(C)-\Pr(A\cap B)-\Pr(A\cap C)-\Pr(B\cap C)+\Pr(A\cap B\cap C).$$

We are told $\Pr(A\cup B\cup C)$, and we computed the next three items in the above formula, so now we know $\Pr(A\cap B\cap C)$.

OK, time to fill in the bits in the Venn Diagram. First write the computed $\Pr(A\cap B\cap C)$ in the space where it should go.

We know $\Pr(A\cap B)$. Since we also know $\Pr(A\cap B\cap C)$, we can find the probability of "the rest of" $A\cap B)$, called $A\cap B\cap C'$ in set language. Compute, write the number in.

Do the same for the other similar bits, namely $A\cap C\cap B'$ and $B\cap C\cap A'$.

Now from $\Pr(A)$ you can find the probability of the "outside" part of $A$, that is, $\Pr(A\cap B'\cap C')$. Do the same for the "outside" part of $B$, also the "outside" part of $C$.

The probability of the outer world $A'\cap B'\cap C'$ of cheapskates who want nothing can now be found in various ways.

We now know everything, and can answer any question!

Remark: You can be more fancy, and let $X$, $Y$, and $Z$ respectively be $A'$, $B'$, and $C'$. By taking every one of the given numbers and subtracting it from $1$, you can write down immediately $\Pr(X)$ (namely $1-0.40$), $\Pr(Y)$, $\Pr(Z)$, $\Pr(X\cap Y)$, and so on. Now we end up with a problem of a type that you have probably done several times.

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So, in the question, when it said something like, "63% request A or B," you assumed that two be $P(A \cup B)=63\%$? Why isn't it really $P(A \cup B \cap C')=63\%$? If you take, "63% request A or B," to mean the former, why wouldn't you include something about C? Because when you write the Venn diagram, A and B both intersect C. – Mack Feb 1 '13 at 14:56
The reason I ask is because $P(A \cup B)$ also includes the probability of getting $A$ and $C$ and $B$ and $C$, and it just seems odd that you wouldn't mention something like that in the question. – Mack Feb 1 '13 at 15:17
@EliMackenzie: The problem setter should make things clear. But I am reasonably sure that the intent here is that "or" be interpreted as $\lor$, logical or, meaning in this case union. There is a somewhat similar potential issue with "and" in standard Venn diagram work problems. When I say $A$ and $B$, does that imply anything about $C$, $D$? No, and you are accustomed to that. – André Nicolas Feb 1 '13 at 16:49
One can also go through the Venn diagram, trying to see whether your interpretation is numerically consistent with the data. The probabilities of $A\lor B$, $B\lor C$, $A\lor C$ are quite large, one can't find a Venn diagram that works. Some of the percentages of "subareas" would turn out negative. – André Nicolas Feb 1 '13 at 16:56
Also, if you look at the text of the problem, it says explicitly as a hint what is meant by $A$ or $B$. That is neutral about $C$. It looks as if the problem setter was worried people might have trouble with the meaning of or. – André Nicolas Feb 1 '13 at 17:03