Solution: 1/6.
Solution using events:
P(both are inoculated) = P(1st in inoculated and 2nd is inoculated) = P(1st is inoculated) P(2nd is inoculated | 1st is inoculated) = (2/6) * (1/5) = 2/30 = 1/15.
Solution using combinations:
We must choose two of two inoculated rabbits and zero of four rabbits that have not been inoculated.
P(both are inoculated) = _{2}C_{2} * _{4}C_{0} / _{6}C_{2}
= (2!/(2!0!)) * (4!/(0!4!)) / (6!/(2!4!)) = 1/15.
Solution using permutations:
With permutations we consider the order in which the rabbits are selected.
P(both are inoculated) = _{2}P_{2} / _{6}P_{2}
= (2*1) / (6*2) = 1/15.
Consider a two by two table which partitions individuals by a positive or negative screening and presence or absence of a disease.
| disease (D) | no disease (~D) | total ----------------------------------------------------------- positive screen (S) | a | b | a+b negative screen (~S)| c | d | c+d ----------------------------------------------------------- total | a+c | b+d | a+b+c+d = N
We define the following terms.
prevalence rate = (a+c)/N = P(D).
sensitivity = a/(a+c) = P(S|D).
specificity = d/(b+d) = P(~S|~D).
false positive rate = b/(b+d) = P(S|~D).
false negative rate = c/(a+c) = P(~S|D).
predictive value = a/(a+b) = P(D|S).
Notice that teh false positive rate is 1 - specificity and the false negative rate is 1 - sensitivity.
Bayes' rule is useful for finding the predictive value or yield.
Solution:
S 0.0095 / 0.95/ / D < / \ / 0.05\ 0.01/ \ / ~S 0.0005 / \ \ S 0.0297 0.99\ / \ 0.03/ \ / ~D < \ 0.97\ \ ~S 0.9603
Thus, P(D|S) = P(D and S) / P(S) = 0.0095 / (0.0095 + 0.0297) = 0.242.
A positive screen (or presence of a symptom) increases the risk of disease from 1% to over 24%. However, it is still true that about three of every four posaitive screens are false, because the disease incidence is so low.
Bret Larget, larget@mathcs.duq.edu