Amazon interview question—probability of 2nd interview I got this question during an interview with Amazon:


*

*50% of all people who receive a first interview receive a second interview

*95% of your friends that got a second interview felt they had a good first interview

*75% of your friends that DID NOT get a second interview felt they had a good first interview


If you feel that you had a good first interview, what is the probability you will receive a second interview?
Can someone please explain how to solve this? I'm having trouble breaking down the word problem into math (the interview is long over now). I understand there may not be an actual numerical solution, but an explanation of how you would walk through this problem would help.
edit: Well I did get a second interview. If anyone is curious I had gone with an explanation that was a combination of a bunch of the responses below: not enough info, friends not representative sample, etc and just talked through some probabilities. The question left me puzzled at the end though, thanks for all of the responses.
 A: *

*State that none of your friends are also up for interview.

*State that the question is underconstrained.


Before they can scramble for some further constraint to the problem quickly try and get in a more productive pre-prepared question of your own in a manner fully expecting a response. Maybe you can get them to move on to a more productive interview.
A: Joke answers but should work well: 


*

*"100% When it comes to demanding superb performance from myself, I don't attribute the outcome to any probability. See you in the 2nd interview."

*"50%, until my friends got their own Amazon Prime account I won't consider their feelings valid. Actually, sorry, that was a bit too harsh. Let me take it back and rephrase: I won't even consider them human beings."

*"Wait, no one ever made my whiny friends feel good. What are your secrets? I want to work for Amazon; give me a chance to please to unpleasable!"

*Fake a phone vibration "Oh, sorry! It was just my Amazon Prime account telling me that the Honda I ordered was shipped. Where were we?"

*"Regardless, I still feel you should send those who didn't get a 2nd interview a 1-month free trial of Amazon Prime. No one should live their life without knowing its glory. And once we got them, retention, retention, retention."

*"55.9% All my friends have an Amazon Prime account and I will make sure to make their experience counts."

A: The question contains insufficient information to answer the question:
$x$% of all people do A
$y$% of your friends do B
Unless we know the population size of all people and your friends, it is not possible to answer this question accurately, unless we make either of two assumptions:


*

*The group your friends is representative of the overall population.  This results in Vincent Galinas' answer or, equivalently, Alex Williams' answer.

*The group your friends is not representative, and is much smaller than the overall population.  This results in CeeJeeB's answer.



Edit: Do also read the comment by Kyle Strand below.  Another aspect we should consider is how similar am I to my friends?.  This depends on whether one interprets you as the person spoken to or as an unspecified individual or group of individuals (both usages exist).
A: Simple case : 
95 / (95 + 75) ≈ 0.559 is a quick way to get to the result  Out of people who felt good - 95 succeeded , 75 failed . So thats probability of you passing from that group is above . But 


*

*No where it is said you are part of the above group . 

*If you can think that distributions (your friends circle's) pattern
is generic or you are in that group you might as well compute this
way 

*Also IMO not that it matters much but the facts about your friend feelings NEED not have any implication in future - that way its worded . For example it rained yday doesn't mean there is a possibility of rain tommorow unless  


Facts , like 50% clearing is not affecting the probability of "what you feel" and the "chances of getting based on that" in that case.
Safer Approach : 
However I even would have thought of the 50% thingy above . I.e from the perspective of real facts - 50% is probability makes sense . 1) No where does it say your feelings SHOULD have anything to do with your results .2) There could be  ppl who are your friends - but HAD NO feelings - what happened to them ... So given all the combinations that are possible - stick with the safest choice !
PS: I might have flunked this test too. 
A: It might be helpful to view this chain of events as a binary tree where just two leaf probabilities are relevant. The root node contains all folks who had a 1st interview; we then split this group on being invited to a 2nd interview ("2nd", "no 2nd") and subsequently on whether they felt good about the 1st interview ("good", "bad").
The conditional probabilities $P(\text{good} | \text{2nd}) = 0.95$ and $P(\text{good} | \text{no 2nd}) = 0.75$ are positioned at the edges. 
For educational reasons, we present two versions: one with decimal probabilities and  the other with absolute counts -assuming a population of $200$.  

We can now directly compute the conditional probability (of receiving a second interview given that you had a good first interview) from the leaf probabilities in two ways:
$$
P(\text{2nd } | \text{ good}) = \frac{P(\text{2nd} \cap \text{good})}{P( \text{good})} = \frac{0.475}{0.475 + 0.375} = \frac{95}{95 + 75} \approx 0.56 
$$
A: The answer is 50%. Particularly since it was an interview question I think Amazon wanted to test the candidate to see if they could spot the obvious and not be distracted by the unimportant.

When you hear hoofbeats, think horses, not zebras - reference

My explanation:  The first statement is all the information you need.
50% of All People who receive first interview receive a second interview

The other two statements are just observations. Feeling you had a good interview does not increase your chances of having a second. 
Although statistically the observations may be correct I believe they cannot be used to predict future outcomes. 
Consider the following.  


*

*2 shops sell lottery scratch cards

*After selling 100 cards each a customer gets a winning card from shop 1

*Statistically you could say that shop 1 now has a greater chance of a person getting a winning ticket, 1 in 100 compared to 0 in 100 for shop 2.


We understand this is not true. The reason it is not true is because in this example past events will not have a bearing on future outcomes.
A: The answer is 50%. They told you in the first line what the chance of anyone getting a second interview is. It's a test of your ability to see the essential information and not get distracted by irrelevant noise like how your friends felt. How they felt made no difference.
A: Both statements say:

% of your friends

not

% of your friends who were interviewed

We do know that the group "that got a second interview" can only include those who had a first interview. However, the group "that did not get a second interview" includes all other friends.
Without knowing what percentage of your friends were interviewed, it is impossible to identify any correlation between feeling you had a good first interview and receiving a second.
A: This being an interview question, I don't believe there is a correct answer.
I would most likely calculate the ~56% using Bayes and then tell the interviewer:
Without any knowledge about me, it could be between 50% and 56%, but because I know me and my past, the probability is 100%
A: Say 200 people took the interview, so that 100 received a 2nd interview and 100 did not. Out of the first lot, 95 felt they had a great first interview. Out of the 2nd lot, 75 felt they had a great first interview. So in total 95 + 75 people felt they had a great first interview. Of those 95 + 75 = 170 people, only 95 actually got a 2nd interview. Thus the probability is:
$$\frac{95}{(95 + 75)}=\frac{95}{170}=\frac{19}{34}$$
Note that, as many commenters graciously point out, this computation is only justifiable if you assume that your friends form an unbiased and well distributed sampling set, which may be a strong assumption.
A: The answer that I would give is:

Based on this information, 50%. 'Your friends' is not a representative sample so it should not be considered in the probability calculation.

If you assume that the data is valid then Bayes' theorem is the way to go.
A: Let


*

*$\text{pass}=$ being invited to a second interview,

*$\text{fail}=$ not being so invited,

*$\text{good}=$ feel good about first interview, and

*$\text{bad}=$ don't feel good about first interview.


$$
\begin{align}
p(\text{pass}) &= 0.5 \\
p(\text{good}\mid\text{pass}) &= 0.95 \\
p(\text{good}\mid\text{fail}) &= 0.75 \\ 
p(\text{pass}\mid\text{good}) &= \;?
\end{align}
$$
Use Bayes' Rule
$$
p(\text{pass}\mid\text{good}) =
 \frac{p(\text{good}\mid\text{pass})
 \times
 p(\text{pass})}{p(\text{good})}
$$
To solve, we need to realize that:
$$
\begin{align}
p(\text{good}) &= p(\text{good}\mid\text{pass})\times p(\text{pass}) + p(\text{good}\mid\text{fail})\times p(\text{fail}) \\&= 0.5(0.95 + 0.75) \\&= 0.85
\end{align}
$$
Thus:
$$
p(\text{pass}\mid\text{good}) = \frac{0.95 \times 0.5}{0.85} \approx 0.559
$$
So feeling good about your interview only makes you slightly more likely to actually move on.
Edit: Based on a large number of comments and additional answers, I feel compelled to state some implicit assumptions. Namely, that your friend group is a representative sample of all interview candidates.
If your friend group is not representative of all interview candidates, but is representative of your performance (i.e. you and your friends fit within the same subset of the population) then your information about your friends could still provide predictive power. Let's say you and your friends are a particularly intelligent bunch, and 75% of you move on to the next interview. Then we can modify the above approach as follows:
$$p(\text{pass}\mid\text{friend})=0.75$$
$$p(\text{good}\mid\text{pass, friend})=0.95$$
$$p(\text{good}\mid\text{fail, friend})=0.75$$
$$
p(\text{pass}\mid\text{good, friend}) =
 \frac{p(\text{good}\mid\text{pass, friend})
 \times
 p(\text{pass}\mid\text{friend})}{p(\text{good}\mid\text{friend})} =
 \frac{0.95 \times 0.75}{0.85} \approx 0.838
$$
A: I think the answer is 50% - right at the beginning of the question. It's irrelevant what percentage of your friends feel. 
A: Mathematically

You're chances are 50%. This is because in the Venn diagram of Amazon Interviewees you fall into the Universal Set of ALL Interviewees, but not the set of 'Your friends'.

Had the question stated: 'One of your friends had a great interview. What is the percentage she'll get a second interview?' Then the current top answer would be valid. But those 2nd and 3rd statistics only apply to you if you consider yourself one of your own friends. So, maybe it's more of a psychological question?
A: Answer is: ≈1
The question doesnt provide how many people among those appearing for interview,are our friends.However, we can assume that data & get any answer we want.Also, main thing about this assumption is that only our friends get selected for 2nd interview.
Lets say 104 of your friends appear for the interview,& 100 of them get 2nd interview.
So, we can say 95 of them felt they had a good first interview(Criteria 2).Also, out of remaining 4,75%(ie 3) of them felt they had a good interview(Criteria 3).So out of 104, 98 felt they had a good interview.but 95 were selected.so final probability is : 95/98.We can always say that 100*2 = 200(104 are friends out of them) people in total gave the first interview, in order to satisfy the 1st criteria.here, all 96 who were not friends,failed to clear 1st interview.
Now you increase friends to 108 & do it again, for 100 of them getting 2nd interview.your final probability would be 101/108 .Thus, as we increase no of friends who didnt clear first interview, the probability decreases.So for maximum efficiency, no of friends who didnt clear should always be 4.
Now increase the friends.Suppose they are 10,004(10000 who cleared,4 who didnt).
so now, out of 10000,9500 felt they had a good interview.So in total, 9503(among 4 failed,3 felt they had good interview, therefore 9500+3) felt they had a good interview,but only 9500 cleared. ie final probability = 9500/9503 which is ≈1.Again, we can put that 20000 people in total appeared for the interview, & all those who werent friends, couldnt clear it.So, 1st criteria is again satisfied.
Note: Our assumption about no of friends,no of them clearing the interview & no of other participants, is all in order to get the probability to 1.we can modify this data & can get any probability we want.
