Find the probability that an employee is a drug user

Question

A firm wants to know how many of its employees have drug problems. Realizing the sensitivity of this issue, the personnel director decides to use a randomized response survey.

Each employee is asked to flip a fair coin,

If head (H), answer the question “Do you carpool to work?”

If tail (T), answer the question “Have you used illegal drugs within the last month?”

Out of 8000 responses, 1420 answered “YES” (assuming honesty)

The company knows that 35% of its employees carpool to work. What is the probability that an employee (chosen at random) used illegal drugs within the last month?

I think the probability that I am trying to figure out is $\mathbb{P}(yes|T)$. From the problem, I was able to figure out that $\mathbb{P}(yes)=1420/8000$, $\mathbb{P}(T)=50%$ (because it's a fair coin) and that $\mathbb{P}(yes|H)=35%$. But for Bayes' theorem, I need to find $\mathbb{P}(T|yes)$, and that is where I am stuck.

I realized that I did not need Bayes' theorem as that would have made it more difficult.

Answer 1

Set up a table like this, with $8000$ people in total:

	$H$	$T$
Yes	$(1)$	$(2)$
No	$(3)$	$(4)$

As you have said, $(1) + (3) = 4000$ people, as the coin is fair. Now the $4000$ people in the 'heads' condition are a representative sample of the company, so $35 \%$ of those carpool as well. This means $(1) = 4000 \times 0.35 = 1400$, and since $1420$ answered 'Yes', $(2) = 1420 - 1400 = 20$:

	$H$	$T$
Yes	$1400$	$20$
No	$4000-1400$	$4000-20$

and therefore the proportion of drug users is $\frac{20}{4000} = \frac{1}{200}$ or $0.005$.

Answer 2

Heuristically:

Assume for convenience that we have $2000$ employees and $1000$ get heads and $1000$ get tails, equally distributed over (non-)carpoolers (of which we have $700$ by the given percentage). Suppose they all answer , we get $\frac{1420}{4} = 355$ yes answers (scaled from $8000$)

Out of the $1000$ heads, we will get $350$ yes answers from carpoolers, and out of the 500 tails we got an extra $5$ yes from drugs users. So we'd expect $\frac{5}{1000}$ as an estimate of the drug users fraction.