# Conditional Probability question - is this info enough to solve it?

This is exercise 3.97 from "Statistics for Business and Economincs", Ed. Newbold, 8th edition (I hope I am allowed to copy it here now that I've given attribution). I am struggling with solving it because I don't think there is enough information.

A company places a rush order for wire of two thicknesses. Consignments of each thickness are to be sent immediately when they are available. Previous experience suggests that the probability is 0.8 that at least one of these consignments will arrive within a week. It is also estimated that, if the thinner wire arrives within a week, the probability is 0.4 that the thicker wire will also arrive within a week. Further, it is estimated that, if the thicker wire arrives within a week, the probability is 0.6 that the thinner wire will also arrive within a week. a. What is the probability that the thicker wire will arrive within a week? b. What is the probability that the thinner wire will arrive within a week? c. What is the probability that both consignments will arrive within a week? Blockquote

Here is where I am with it: Let T1= thin wire arrives within a week; T2= thick wire arrives within a week; P(T1UT2) = 0.8; P(T2|T1) = 0.4; P(T1|T2) = 0.6 We are looking for: a) P(T1); b) P(T2); c) P(T1∩T2).

It seems to me, however, that this is not enough; that I would need either the intersection, or one of the individual probabilities. What am I missing? Or misunderstanding?

• Draw the Venn diagram. You are given three facts about the probabilities of the four regions in it and you know a fourth fact automatically (they sum to unity). Solve these equations.
– whuber
Feb 13 '20 at 20:19
• @whuber, I can't draw the Venn diagram because two of the values given are conditional probabilities, which adds a time-factor, or a 3rd dimension, which I can't draw in 2-d. I could if this was an P(AUB) = P(A) + P(B) - P(A∩B) type of question - but it isn't. Which is why I am struggling. :-( Feb 13 '20 at 21:09
• There's no third dimension. When you are told $\Pr(A \mid B) = 0.4,$ for instance, that's equivalent (according to the definition of conditional probability) to $0.4 = \frac{\Pr(A \cap B)}{\Pr(B)},$ implying $0.4 \Pr(B) = \Pr(A\cap B).$ That's a linear equation between two of the regions in the Venn diagram. You will have no problems solving this system of linear equations.
– whuber
Feb 13 '20 at 21:36
• So that gives me 0.6T2=0.4T1, didn't get me far (I know the proportions but they can still be anything); also 0.6= 0.6T2/T2, in other words 0.6=0.6, not helpful either. If I take the union, I get T1= 0.8+1.6T2, which then gives T1|T1=0.6T2/0.8+1.6T2, which is obviously nonsense, given 0.4=/=0.25 ... (Also, this is a very basic chapter in an introductory book; I have the solution manual for all the even-numbered exercises in the chapter and none of those require anything like this, they are all just basic applications of the rules, so there has to be a simple solution.) Feb 17 '20 at 11:50

Take the following ratio: $${{P(T_1|T_2)}\over{P(T_2|T_1)}}={P(T_1)\over P(T_2)}={3\over2}$$ You can write (let $$P(T_1)=x$$) $$P(T_2),P(T_1\cap T_2)$$ in terms of $$x$$.
Also, you have $$P(T_1\cup T_2)=P(T_1)+P(T_2)-P(T_1\cap T_2)$$ in which you only have $$x$$ as the unknown.
• Write as $$P(T_1|T_2)=\frac{P(T_1\cap T_2)}{P(T_2)}$$ Feb 13 '20 at 21:29
• @Reader123 Let me write it more explicitly: $$\frac{P(T_1|T_2)}{P(T_2|T_1)}=\frac{\frac{P(T_1\cap T_2)}{P(T_2)}}{\frac{P(T_1\cap T_2)}{P(T_1)}}=\frac{P(T_1)}{P(T_2)}$$ Feb 17 '20 at 11:57