# Are you Really Smart

The government has introduced a standardised test to identify dumb people and make them ineligible to apply for any government jobs, a pass in the exam is mandatory for applications in future.

It was hard to come up with a test that can provide 100% accuracy, however the creators of the test argued that 289 of every 300 dumb people taking the test will fail and 195 of every 200 smart people taking the exam would pass it and since it left very little to desire, the test was implemented. Given the general nature of people applying for such jobs, it was predicted that 60% of the smart people would not take this exam and all of the dumb people will take it, leaving the percentage of smart people taking the exam at 40% (To avoid confusion mentioned in the answers this means of every 100 people taking the exam 40 were smart and 60 were dumb)

You took the test and you passed, what is the probability that you are really smart?

• What is the fraction of smart people in the general population? This question is unanswerable without that (or equivalent) information. Oct 24 '14 at 13:37
• @frodoskywalker I thought that could be another question we could add, or am I missing it because I saw it so many times while creating it
– skv
Oct 24 '14 at 13:51
• so you're saying that people can optout of taking it. sorry I don't get how 60% of smart could simply not take the test Oct 24 '14 at 13:57
• @d'alar'cop yes 60% of the smart population opted out, and that left the overall number of dumb people taking the exam higher (edited the question)
– skv
Oct 24 '14 at 13:59
• I know this puzzle in much nicer formulation: "0.1% of people are genius. There is a test to distinguish genius people and others, the error of test is 1%. I am genius according to the test, what is probability that I am usual?" In this case it's really a puzzle, it is worth to calculate the answer, and people tend to make mistakes. I wonder should I make a separate question from this.. Oct 24 '14 at 14:22

I am really smart so 100%.

If Y is the fraction of people who take the test that are smart then $YT*195/200$ smart people will pass the test where $T$ is the number of test takers. $(1-Y)T*11/300$ is the number of dumb people who will pass. The percentage of people who will pass that are smart is: $$\frac{YT*195/200}{YT*195/200+(1-Y)T*11/300}=\frac{585Y}{563Y+22}$$

If $Y =\frac{40}{100}$ then $$\frac{585Y}{563Y+22}=94.66\%$$

If we want to generalise this further to answer klm123's version of the question, I would define $A$ as the fraction of geniuses who pass and $B$ as the fraction of idiots who pass. (Sorry but you are one or the other according to this.) The proportion of people who pass who are geniuses are, therefore:

$$\frac{AYT}{AYT+B(1-Y)T}=\frac{AY}{(A-B)Y+B}$$

For $Y=.1\%$, $A=99\%$, and $B=1\%$ this is roughly only $9\%$.

• @klm123 "this means of every 100 people taking the exam 40 were smart". Did I misinterpret this? Oct 24 '14 at 14:19
• @klm123 are you looking at the updated question? It was edited to include the quote in teh previous comment. Oct 24 '14 at 14:26
• Damn, I am bad a arithmetic today...the equation is right though! Oct 24 '14 at 17:01

Say $3000$ people applied.

$1200$ of these are smart ($40\%$), and $1800$ are dumb ($60\%$).

$(1200\frac {195}{200})+(1800\frac {11}{300}) = 1236$ people out of $3000$ applicants passed.

It is known that of the $1236$, $(1200\frac {195}{200}) = 1170$ people are smart and $(1800\frac {11}{300})=66$ are dumb by the probabilities passing/failing provided.

So, there is a $\frac{1170}{1236} = 0.94660194174$ probability that you are smart.

• I really would like to select yours because it is my way of solving things (taking a number instead of Alphabets) but since it came in a minute later (the edits) I had chosen the other answer
– skv
Oct 24 '14 at 14:11

Just thought I would write an answer using normal probability notations and finish by plugging in the values...

• $P$ = Passing,
• $D$ = Being Dumb,
• $S$ = Being Smart

We want to get p(S|P)

$p(S|P) = \frac{p(S\&P)}{p(P)}$ $= \frac{p(P|S)p(S)}{p(P)}$ $= \frac{p(P|S)p(S)}{[p(P|S)p(S)+p(P|D)p(D)]}$

plugging in...

$=(195/200)(40/100)/[(195/200)(40/100)+(11/300)(60/100)]$

$=94.66%$

In fact, I did not pass the test, because this test about detecting smart people makes a mistake at the very beginning. So... Instead of passing it, I came and proposed a correction :

$289$ out of $300$ true negatives and $195$ out of $200$ true positives is not $100\%$ accuracy. It's only $96.8\%$ accuracy.

• where in the question do you feel its indicated as 100%
– skv
Oct 25 '14 at 7:59
• This is at best a comment to the question not an "Answer"
– skv
Oct 25 '14 at 8:00
• Indeed, he only says it was hard to come up with a test that can provide $100\%$ accuracy. It does not says explicitly, but one could interpret it this way. Oct 25 '14 at 8:02
• Its my question, if you read on... I have mentioned clearly that "since it left very little to desire, the test was implemented" which makes it very clear that they did not succeed in coming up with a test that could provide 100% accuracy
– skv
Oct 25 '14 at 8:03
• My bad then. But you probably should edit your question. Say something like : "Given the state of our knowledge, it was not possible to come up with a test that provide $100\%$ accuracy...". Oct 25 '14 at 8:08