Knowing that the first test was positive, what is the probability the second test will be positive? Suppose that 5% of the employees of a certain company use illegal drugs. The company performs random drug tests that return positive results 98% of the time if the person is a drug user. However, it also has a 4% false-positive rate. The results of the test are independent from test to test for a given person.

*

*An employee at the company has a positive test. What is the probability that he uses drugs?


*Knowing that the first test was positive, what is the probability the second test will be positive?


*Suppose that the second test is also positive. What is the probability he uses drugs?
My Attempt

*

*This one is straightforward.. From the problem, we know
$$\mathbb{P}(positive|\;uses\;drugs)=0.98$$
$$\mathbb{P}(uses\;drugs)=0.05$$
$$\mathbb{P}(positive|doesn't\;use\;drugs)=0.04$$
$$\mathbb{P}(doesn't\;use\;drugs)=0.95$$
Then using Baye's Theorem, we have
$$\mathbb{P}(Uses\;drugs)=\frac{(0.98)(0.05)}{(0.98)(0.05)+(0.04)(0.95)}=0.56$$


*For this one, we are trying to find $\mathbb{P}(Second\;positive|first\;positive)$. Since the results are independent from each other, we know that $\mathbb{P}(Second\;positive|first\;positive)=\mathbb{P}(second\;positive)$. Would this mean that the second test being positive has the same probability as the first one? Intuitively, I would think that the probability is the same because the events are independent of each other, but I am not sure how to prove this.

Edit:
I think I can use the definition of conditional probability and the law of total probability to solve this problem. What I had above is wrong.
I can use the following to solve:
$$\mathbb{P}(B\;|\;A)= \frac{\mathbb{P}(B\cap A)}{\mathbb{P}(A)}= \frac{\mathbb{P}(B\cap A\;|\;D)\mathbb{P}(D)+\mathbb{P}(B\cap A\;|\;D^c)\mathbb{P}(D^c)}{\mathbb{P}(A\;|\;D)\mathbb{P}(D)+\mathbb{P}(A\;|\;D^c)\mathbb{P}(D^c)}$$


*For this one, I am trying to find $\mathbb{P}(Uses\;drugs|first\;positive\;\cap\;second\; positive)$. I think I can use Baye's rule for this one again and the definition of conditional independence.

 A: You nailed the first test, which means that the chance that the employee is a drug user are no longer $(0.05)$ but have changed to 
$\displaystyle \frac{49}{87}.$
Therefore, you attack the second question in the exact same way that you attacked the first question, except that the probability has changed.
Killing two birds with one stone, consider the following computation, which relates to the second test.
$$\frac{(49/87) \times (0.98)}{[(49/87) \times (0.98)] + [(38/87) \times (0.04)]}.$$
The chance of the second test being positive is represented by the denominator in the above fraction.  The chance of the employee being a drug user, after two consecutive positive results is represented by the above fraction.
A: For 2 and 3,  the independence just means that the probability that the second test is a false positive is stlll 4%. It does not mean that the event of a positive test on round 2 is independent of the event of a positive test on round one (which, as you say, would be extremely counterintuitive).
