0
$\begingroup$

I'm trying to understand conditional probability and I thought of this problem, I'm not sure it might be related.

I lent 500$ to a friend of mine, he told me that there are 20% chances that he gives me the money back in August and 70% chances that he pays me back in September.

I labelled the event of getting paid in August as $A$ and getting paid in September as $B$, therefore $P(A) = 0.2$ and $P(B) = 0.7$.

The question is: we move forward to the end of August, my friend could not pay me back then, what are the chances that I get paid in September?

Basically I want to know $P(B\,|\,\overline{A})$ where $\overline{A}$ is the event of not getting paid in August, therefore I want to know what are the chances of getting paid in September (i.e. $B$) given that I was not paid in August (i.e.: $\overline{A}$).

Calculating the conditional probability using the Bayes' rule I get: $$ P(B\,|\,\overline{A}) = \frac{P(\overline{A}\,|\,B)P(B)}{P(\overline{A})} $$

where

$P(\overline{A}\,|\,B) = 1.0$ because if I get paid in September there is a 100% probability that I was not paid in August

$P(B) = 0.7$

$P(\overline{A}) = 1-0.2 = 0.8$

Therefore the above calculation becomes $\frac{P(B)}{P(\overline{A})}$ or $\frac{0.7}{0.8} = 0.875$

Hence there is a 87.5% chance that I will get paid in September given that I was not paid in August. Is it correct?

In a variation of the problem, my friend tells me that if he cannot pay me in September there will be a 30% chance that he will pay me in October, in that case will I recalculate the conditional probability using (1-0.875) as the new prior? Is it a correct approach?

$\endgroup$
  • $\begingroup$ What conditional are you interested in computing? You already have the probability of getting the money in October if your friend does not pay in September. $\endgroup$ – broncoAbierto Jul 23 '16 at 17:55
  • $\begingroup$ are you referring to the second part of the problem? Is the first part correct? The second part was written under the assumption that at the beginning of the problem I'm given probabilities for 3 months rather than 2 $\endgroup$ – complexguest Jul 23 '16 at 18:03
  • $\begingroup$ Yes to both questions. $\endgroup$ – broncoAbierto Jul 23 '16 at 18:06
  • $\begingroup$ So if the initial probabilities are: August (0.2), September (0.7), October (0.3), if I'm not paid in August, then the September probability becomes 0.875 according to the first part of the problem, so I suppose that if I'm not paid in September the Oct probability has to change too $\endgroup$ – complexguest Jul 23 '16 at 18:10
  • $\begingroup$ Are you sure that initial setting is possible? $\endgroup$ – broncoAbierto Jul 23 '16 at 18:17
1
$\begingroup$

By popular request, I am reposting my comments as an answer.

The way you wrote it originally, $0.3$ probability of being paid in October is (already) conditional on not being paid in September. It (0.3) can not be the unconditional probability of being paid in October, because this would imply the unconditional probability of being paid is at least $1.2 = 0.2 + 0.7 + 0.3$. Therefore, no further adjustment is in order to get the conditional probability of being paid in October given not being paid in September.

On the other hand, the $0.7$ probability of being paid in September was stated as being an unconditional probability; hence your calculation to get the conditional probability of being paid in September given not being paid in August.

Your calculation to determine the probability of being paid in September, given not being paid in August, is correct. As a partial check of your calculation, note that if the unconditional probabilities of being paid in August and September were $0.2$ and $0.8$ respectively, rather than $0.2$ and $0.7$, the corresponding calculation to what you did would result in a probability of being paid in September, given not being paid in August, of 1, which of course is as it must be given the unconditional probability of being paid in August or September being equal to $1$ .

As a practical note, if this is just a hypothetical (theoretical) exercise, fine ("in a variation of the problem" doesn't sound like the way you'd describe a real-world situation). But if this is your real-life situation, I don't trust the probabilities provided by your friend for a variety of reasons, even if he is basically honest. If you don't get paid by the end of September, I think you can kiss your money and friendship goodbye.

$\endgroup$
0
$\begingroup$

The question was answered in the comment section by Mark L. Stone and broncoAbierto.

From broncoAbierto:

The first part is correct. You can interpret it intuitively the following way. The chances of getting paid in September have a 7 to 1 ratio with respect to the chances of getting paid any other time except August (because $P(B)=0.7$). Once you rule out August, the probability of getting paid in September is 0.875, which is seven times 0.125, (the probability of getting paid later).

From Mark L. Stone:

The way you wrote it originally, 0.3 probability of being paid in October is (already) conditional on not being paid in September. It can not be unconditional probability of being paid in October, because this would imply at least a 1.2 = 0.2 + 0.7 + 0.3 unconditional probability of being paid. Whereas 0.7 probability of being paid in September was stated as being an unconditional probability; hence your calculation to get the conditional probability of being paid in September given not being paid in August. If you don't get paid by September, you can kiss your money and friendship goodbye."

I'm happy to remove this answer and accept either answers if one of them reposts their comments as answer.

$\endgroup$
  • 2
    $\begingroup$ Comments are "second class citizens" of Stack Exchange - they could theoretically be wiped out at any time! So it would be better to include some details about what those comments said in your answer. (It is fine to quote someone else's comment in an answer.) $\endgroup$ – Silverfish Jul 23 '16 at 19:21
  • $\begingroup$ Thanks for editing your answer to include the comments themselves. We actually have formatting for quotes - just put > in front of the paragraph. $\endgroup$ – Silverfish Jul 23 '16 at 20:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.