Consider influenza epidemics for two parent heterosexual families. Suppose that the probability is 17% that at least one of the parents has contracted the disease. The probability that the father has contracted influenza is 12% while the probability that both the mother and father have contracted the disease is 6%. What is the probability that the mother has contracted influenza ?
Answer : Let call A the probability that the father contracted the disease and B the probability that the mother contracted the disease. We have \[{P(A \cup B) = 0.17}\] \[{P(A) = 0.12}\] \[{P(A \cap B) = 0.06}\]
We also know that \[{P(A \cup B) = 0.17 = P(A) + P(B) - P(A \cap B)}\]
So P(B) = 11%.
A random variable X is uniform, a box from 0 to 1 of height 1. (So that its density is \({f(x) = 1 for 0\leq x \leq 1}\)). What is its 75% percentile ? (Hint : look at lecture 2 at 21:30 and chapter 5 problem 5. Also look up the help function for the qunif function in R).
Answer : As the density is constant, the point that the area below is 0.75 is also 0.75.
You are playing a game with a friend where you flip a coin and if it comes up heads you give her X dollars and if it comes up tails she gives you Y dollars. The probability that the coin is head is \({p}\) (some number between 0 and 1). What has to be true between X and Y to make so that both of your total earnings is 0 ? The game would then be “fair”.
Answer : We must have \({-p.X + (1-p).Y = 0}\). So \({\frac{p}{1-p} = \frac{Y}{X}}\).
A density that looks like a normal density (but may or may not be exactly normal) is exactly symmetric about 0. (Symmetric means if you flip it around 0 it looks the same). What is the median ?
Answer : The median must be 0 because 50% of the mass is below 0 and 50% above.
Consider the following PMF shown below in R :
x <- 1:4
p <- x/sum(x)
temp <- rbind(x,p)
rownames(temp) <- c("X","prob")
temp## [,1] [,2] [,3] [,4]
## X 1.0 2.0 3.0 4.0
## prob 0.1 0.2 0.3 0.4What is the mean ?
Answer :
sum(p*x)## [1] 3So the mean is 3.
A web site (http://medicine.ox.ac.uk/bandolier/band64/b64-7.html) for home pregancy tests cites the following :“When the subjects using the test were women who collected and tested their own samples, the overall sensitivity was 75%. Specificity as also low, in the range 52% to 75%.” Assume the lower value for the specificity. Suppose a subject has a positive test and that 30% of women taking pregnancy tests are actually pregnant. What number is closest to the probability of pregnancy given a positive test ?
Answer : We know that \[{sensitivity = Pr(+|Preg) = 0.75}\] \[{specificity = Pr(-|Preg^{C}) between 0.52 and 0.75}\] \[{Pr(Preg) =0.3}\]
Wee want to compute : \[{P(Preg |+) = \frac{P(+|Preg).P(Preg)}{P(+|Preg).P(Preg) + P(+|{Preg}^C).P({Preg}^C)}}\]
So \[{P(Preg |+) = \frac{0.75*0.3}{0.75*0.3 + (1-0.52)*0.7}}\]
The probability of pregnancy given a positive test is around 40%.