Probability that all $14$ coronavirus tests are correct

Question

A coronavirus test has

• sensitivity of $97\%$, i.e., an infected person is $97\%$ likely to get a positive test.

• specificity of also $97\%$, i.e., a non-infected person is $97\%$ likely to get a negative test.

Question: If $14$ students were tested negative, what is the probability that all of these tests are correct?

Can this question be answered directly, or does it require Bayes' theorem (and the prevalence of the virus in the population). Thanks in advance for answering! :)

Answer 1

Bayes' Theorem says that:

$$\text{P(all tests correct|14N)}=\frac{\text{P(14N|all tests correct)}\text{P(all tests correct)}}{\text{P(14N)}}$$

Assume that $p$ is the chance of infection and that $k$ students are infected:

$$\text{P(14N|all tests correct)}=(1-p)^{14}$$

$$\text{P(all tests correct)}=(0.97)^{14}\approx0.653$$

$$\text{P(14N)}=(0.03)^k(0.97)^{14-k}$$

So the final answer is:

$$\text{P(all tests correct|14N)}=(0.97(1-p))^{14}\sum_{k=0}^{14} \frac{1}{(0.03)^k(0.97)^{14-k}}$$