2.1.11 problem 13

(a)

Let $B$ be the event that the test done by company $B$ is successfull. Let $A$ be the event that the test done by company $A$ is successfull. Let $D$ be the event that a random person has the disease.

$$\begin{array}{cccc}P\left(B\right)& =P\left(D\right)P\left(B\right|D)+P(D^c\left)P\right(B\left|D^c\right)& & \\ & =0.01\ast 0+0.99\ast 1& & \\ & =0.99& & \end{array}$$

$$\begin{array}{cccc}P\left(A\right)& =P\left(D\right)P\left(A\right|D)+P(D^c\left)P\right(A\left|D^c\right)& & \\ & =0.01\ast 0.95+0.99\ast 0.95& & \\ & =0.95& & \end{array}$$

Thus, $P\left(B\right)>P\left(A\right)$.

(b)

Since the disease is so rare, most people don’t have it. Company $B$ diagnoses them correctly every time. However, in the rare cases when a person has the disease, company $B$ fails to diagnose them correctly. Company $A$ however shows a very good probability of an accurate diagnoses for afflicted patients.

(c)

If the test conducted by company $A$ has equal specifity and sensitivity, then it’s accuracy surpasses that of company $B$’s test if the specifity and the sensitivity are larger than $0.99$. If company $A$ manages to achieve a specifity of $1$, then any positive sensitivity will result in a more accurate test. If company $A$ achieves a sensitivity of $1$, it still requires a specificity larger than $0.98$, since positive cases are so rare.