# Bayes with extra conditioning

I have doubt in following proble. Can any 1 please tell whether i did it correctly or not.

A patient named Fred is tested for a disease called conditionitis, a medical condition that afflicts 1% of the population. The test result is positive, i.e., the test claims that Fred has the disease. Fred, who tested positive for conditionitis as above, decides to get tested a second time. The new test is independent of the original test (given his disease status) and has the same sensitivity and specificity. Unfortunately for Fred, he tests positive a second time. Find the probability that Fred has the disease, given the evidence in conditioning on both test results

(Although this is already mentioned in Finding the probability that someone has the disease, given they test positive on two tests ) i have a doubt here. I have gone through all of the post but could not find where i have doubt ... kindly have a look below)

My approach:

D = Fred has disease; $$T_1 =$$ first test result positive ;

$$T_2=$$ second test result positive $$P(D/T_1, T_2) =\frac{ P(T_1,T_2/D)*P(D)}{P(T_1,T_2)}$$

Here "," denotes "intersection"

Here in problem, it is mentioned that test results are independent to each other. SO can i take here $$P(T_1 \cap T_2)=P(T_1)*P(T_2) = 0.059^2=0.003481$$ -->(1)

[Due to this small fraction i am getting $$P(D/T_1, T_2) > 2$$ which is nonsense. It means $$T_1, T_2$$ not independent but problem and common sense says they should be independent. Somewhere i feel i am thinking wrongly

But we can also calculate $$P(T_1 \cap T_2) = P(D)P(T_1 \cap T_2/D)+P(D^c)P(T_1 \cap T_2/D^c) \\ = 0.01*0.95^2+0.99*0.05^2 \\ = 0.0115$$ -->(2)

This is giving me correct answer as given in textbook. Please tell me where i made mistake.

• Sir please have a look now ... it's typing mistake in latex. Commented Oct 12, 2019 at 23:52
• My Question - Is $P(T_1,T_2)=P(T_1)*P(T_2)$ as T_1, T_2 are independent as mentioned in given problem. But this is giving me probability >2 . Commented Oct 12, 2019 at 23:54
• Since $P(T_1,T_2/D)\le P(T_1,T_2)$ and $P(D)\le1$, you can't have $\frac{ P(T_1,T_2/D)P(D)}{P(T_1,T_2)}>1$. Your assumption of independence does not hold: in "The new test is independent of the original test (given his disease status)", the parenthesis means something ("given" $=$ conditioning). But I admit it's not entirely obvious from the problem statement. Commented Oct 12, 2019 at 23:56
• @Jean-ClaudeArbaut thank you sir Commented Oct 13, 2019 at 12:36