Each of 500 soldiers in an army company independently has a certain disease with probability 1/103. This disease will show up in a blood test, and to facilitate matters, blood samples from all 500 soldiers are pooled and tested.

(a) What is the (approximate) probability that the blood test will be positive (that is, at least one person has the disease)? Suppose now that the blood test yields a positive result.

(b) What is the probability, under this circumstance, that more than one person has the disease? Now, suppose one of the 500 people is Jones, who knows that he has the disease.

(c) What does Jones think is the probability that more than one person has the disease? Because the pooled test was positive, the authorities have decided to test each individual separately. The first i − 1 of these tests were negative, and the ith one—which was on Jones—was positive.

(d) Given the preceding scenario, what is the probability, as a function of i, that any of the remaining people have the disease?

Each of $500$soldiers in an army company independently has a certain disease with probability $1/103$. This disease will show up in a blood test, and to facilitate matters, blood samples from all $500$soldiers are pooled and tested.

Since the sample of soldiers is very large, $n=500$and the probability of the disease for every soldier is relatively low, $p={10}^{-3}$, we can use Poisson approximation. Hence, if we mark $X$as the random variable that marks the number of diseased soldiers, we have that $X$has approximatively $Pois\left(np\right)=Pois(0.5)$distribution.

Thus,

$P(X\ge 1)=1-P(X=0)=1-e^{-0.5}\approx 0.393$.

So the probability that at least one soldier will be tested positive is$0.393$.

Each of $500$ soldiers in an army company independently has a certain disease with probability $1/103$. This disease will show up in a blood test, and to facilitate matters, blood samples from all $500$ soldiers are pooled and tested.

Now, we are given the information that the test is positive, i.e. that $X\ge 1$. Now, we have to find the probability of $X>1$given that information.

We have that

$P(X>1\mid X\ge 1)$

$=\frac{P(X>1,X\ge 1)}{P(X\ge 1)}$

$=\frac{P(X\ge 2)}{P(X\ge 1)}$

$=\frac{1-e^{-0.5}-0.5e^{-0.5}}{1-e^{-0.5}}$

$=0.299$.

The probability, under this circumstance, that more than one person has the disease is$0.299$.

Each of$500$ soldiers in an army company independently has a certain disease with probability $1/103$. This disease will show up in a blood test, and to facilitate matters, blood samples from all $500$ soldiers are pooled and tested.

We understand that Jonas carries the disease. But, we accomplish not hold any knowledge concerning any of the other soldiers. Hence, we can exclude Jonas from the story and just assume the remaining $499$ soldiers. But, observe that we want to find the probability that there is more than one soldier with disease and we know that Jonas has the disease, so, we in fact desire to find the probability that some of the remaining $499$ soldiers carry the disease.

So, here repeats story from the part $\left(a\right)$, and since $n$ has been changed just a bit, the required probability also does not change a lot. The answer is $0.393$.

Each of $500$ soldiers in an army company independently has a certain disease with probability $1/103$. This disease will show up in a blood test, and to facilitate matters, blood samples from all $500$ soldiers are pooled and tested.

Do the identical item as in $\left(c\right)$, exclude Jonas and the first $i-1$people from the story. Use $Y$as the random variable that marks the number of a deceased soldier from the remaining $500-i$soldiers. Similarly, $Y~Pois\left((500-i)·{10}^{-3}\right)$.

We want to find the probability that $Y\ge 1$.

Hence

$P(Y\ge 1)=1-P(Y=0)=1-e^{-(500-i)·{10}^{-3}}$.

The probability, as a function of$i$is found to be$1-e^{-(500-i)·{10}^{-3}}$.