A) 8/27

B) 26/27

C) 8/9

D) 1/27

Explain your reasoning along with your answer

(http://www.autoadmit.com/thread.php?thread_id=4181520&forum_id=2#37623215)

🔑

(http://www.autoadmit.com/thread.php?thread_id=4181520&forum_id=2#37624423)

i thought it was a gmat question because op is JJC

(http://www.autoadmit.com/thread.php?thread_id=4181520&forum_id=2#37624468)

Easier to do it backwards and say what are the odds that all three are lying.

1/3 * 1/3 * 1/3 = 1/27

(http://www.autoadmit.com/thread.php?thread_id=4181520&forum_id=2#37623272)

You need a prior distribution to compute it. Think about the intended solution and you’ll see there’s an implicit (and implausible) prior distribution assumed.

(http://www.autoadmit.com/thread.php?thread_id=4181520&forum_id=2#37624290)

Edit: There's 27 ways (combinations) they could answer the question.

They all said "yes". This can only happen if they're all telling the truth, or all lying.

There's 8/27 ways they could all be telling the truth (2/3 to the power of 3).

There's 1/27 ways they could all be lying (1/3 to the power of 3).

So of the 9 ways they could have all answered "yes", 8 have them telling the truth and 1 has them lying, therefore 8/9 chance of telling the truth.

(http://www.autoadmit.com/thread.php?thread_id=4181520&forum_id=2#37623406)

I think the answer really depends on the phrasing of the question. and given that all 3 said "YES its raining" the probability that it is indeed raining is 1-probability they all lied. so it is 26/27

after thinking a little bit more i think 8/9 would be the answer if the question was phrased as "given that all 3 replied the same way what is the probability that it is raining". but the question asked here is different and explicitly says that all said YES

(http://www.autoadmit.com/thread.php?thread_id=4181520&forum_id=2#37624152)

You know that they’re either all lying (it’s not raining) or all telling the truth (it is raining).

Odds that they’re all lying are 1/27 (1/3^3).

Odds that they’re all telling the truth are 8/27 (2/3^3).

The other 18/27 possible states don’t matter.

So, 8 times out of 9, it’s raining.

(http://www.autoadmit.com/thread.php?thread_id=4181520&forum_id=2#37624655)

Fuck, now I see it depends on the base rate.

I haven’t thought about bayes rule in like 10 years.

(http://www.autoadmit.com/thread.php?thread_id=4181520&forum_id=2#37624672)

The only way it is NOT RAINIING is the 1/27 chance ALL OF THEM are fucking with you

Its like saying there you have three radars across Seattle. The radars are accurate 2 out of three times. You check the radars and they all indicate that it's raining. What is the probability it is raining in Seattle (in one part, the whole city, it doesn't matter)?

For it to be NOT RAINING all three radars have to be wrong.

Math major and this took me less than two seconds to think through op

(http://www.autoadmit.com/thread.php?thread_id=4181520&forum_id=2#37626940)

Can you even explain your reasoning?

I can't even believe this thread made it this far. This is literally some high school level shit

Edit: Im pretty sure the part that trips you and other law Boys up is that only one friend needs to be telling the truth, therefore, all of them have to be lying for it to be not raining

(http://www.autoadmit.com/thread.php?thread_id=4181520&forum_id=2#37627022)

There are two possibilities: it’s raining and they’re all telling the truth, or it’s not raining and they’re all lying. You must account for both.

Jfc and don’t you purportedly work in finance?

(http://www.autoadmit.com/thread.php?thread_id=4181520&forum_id=2#37627262)

As OP said, this is a very simple probability question. You've shitted up the whole thread citing irrelevant shit and have somehow convinced yourself that the answer key is wrong along the way lol

What you said makes no sense

(http://www.autoadmit.com/thread.php?thread_id=4181520&forum_id=2#37627501)

Do you agree the answer we seek is the probability of it raining conditional on all three friends saying it’s raining?

(http://www.autoadmit.com/thread.php?thread_id=4181520&forum_id=2#37627571)

I've given you the benefit of the doubt and treated you with wayyyyy too much respect thus far. Lmao I am officially OUT of this thread now. Waste of time

(http://www.autoadmit.com/thread.php?thread_id=4181520&forum_id=2#37627618)

You are computing some probability conditional on some given information (three friends say it’s raining).

(http://www.autoadmit.com/thread.php?thread_id=4181520&forum_id=2#37627627)

"Each of the above analyses fail to assess the logic of the question primarily in their approach. Since all three say it is raining, you cannot approach the solution by determining each friend's propensity for truth. For it not to be raining, they ALL must be lying. Therefore, the solution must be the inverse of the probability that all three are "messing with you." (1/3)x(1/3)x(1/3)=1/27 (3.7% chance they are all lying). Since there is only a 3.7% chance all three friends are messing with you, there is a 96.3% chance it is raining."

This is a great interview question because someone like me will be able to quickly determine the goal of the question, map my way to an answer and walk the interviewer through my process, and then explain my answer to contemplate any potential problems or shortcomings of my answer. Meanwhile, someone like you will be splerging out and flailing their arms about some irrelevant shit, scribbling down numbers and gook hyroglypbics on your paper, before telling the interviewer the Q can be answered after 15 minutes. Then you go home an admire your genius for "winning" the trick question

(http://www.autoadmit.com/thread.php?thread_id=4181520&forum_id=2#37627727)

The issue you’re missing is that the probability of them lying a priori is not the probability of their utterances being false after you’ve been given the information all three say it’s raining. Think about a situation where it never rains in Seattle, for example.

(http://www.autoadmit.com/thread.php?thread_id=4181520&forum_id=2#37627783)

Given they all said it's raining, the range of possible situations changes from 27 to 9, and in 8 of them Seattle is raining.

(This all assumes the probability of raining/not raining is 50/50)

(http://www.autoadmit.com/thread.php?thread_id=4181520&forum_id=2#37629051)

Author: Mormon missionary (only posts on WSHH and xo)

Jfc lawboys

The only way it is NOT RAINIING is the 1/27 chance ALL OF THEM are fucking with you

Its like saying there you have three radars across Seattle. The radars are accurate 2 out of three times. You check the radars and they all indicate that it's raining. What is the probability it is raining in Seattle (in one part, the whole city, it doesn't matter)?

For it to be NOT RAINING all three radars have to be wrong.

Math major and this took me less than two seconds to think through op

(http://www.autoadmit.com/thread.php?thread_id=4181520&forum_id=2#37626983)

http://allendowney.blogspot.com/2016/09/bayess-theorem-is-not-optional.html?m=1

(http://www.autoadmit.com/thread.php?thread_id=4181520&forum_id=2#37627942)