In class problems – some answers
January 15, 2013
1) Of 100 people, 50 are Democrats, 40 are Republicans, and 10 are Independents. The
percentages of the people in these three categories who read The Wall Street Journal are known
to be 30, 60, and 40 percent, respectively. If one of these people is observed reading the Journal,
what is the probability that he or she is a Republican?
Let B represent the event that a person reads the Wall Street Journal and let A1, A2 and A3
denote, respectively, the events a Democrat, a Republican, and an Independent. We want:
P(A2|B) = (0.6)(0.4) / [(0.6)(0.4) + (0.3)(0.5) + (0.4)(0.1)] = 0.558
Note that the prior probability of sampling a Republican is 0.4 or 40%. But knowing that
someone is found reading the Wall Street Journal, the probability of sampling a Republican
increases to 55.8% percent.
2) Let A denote the event that a person lives in New York City. Let P(A) = 0.5. Let B denote the
event that the person does not live in New York City but works in the city. Let P(B) = 0.4. What
is the probability that the person either lives in the city or does not live in the city but works there?
This is P(A+B) = 0.9
3) The following table gives the joint probability of X and Y, where X represents a person’s
poverty status, and Y represents the person’s race:
X
Y
White
Black
Hispanic
Asian
Below Poverty Line
0.0546
0.0315
0.0337
0.0046
Above Poverty Line
0.6153
0.0969
0.1228
0.0406
a) Compute f(X|Y = white); f(X|Y = black); f(X|Y = Hispanic), and f(X|Y = Asian). What
general conclusions can you draw from these computations?
f(X|Y = white) = f(X,Y=white)/f(Y=white) = 0.0546/0.6699 = 0.0815
f(X|Y = black) =0.0315/0.1284= 0.2453
f(X|Y = Hispanic) = 0.0337/0.1565 = 0.2153
f(X|Y = Asian) = 0.0046/0.0452 = 0.1018
b) Are race and poverty status independent variables? How do you know?
They are not. For them to be independent, f(X,Y) = f(X)f(Y) must hold true for all combinations
of X and Y values, which is not the case here.
4) The following table gives joint probabilities relating cell phone usage to stopping properly at
intersections.
On cell phone
Not using cell phone
Failed to stop at intersection
0.047
0.201
Stopped at intersection
0.016
0.736
a) Compute the probability of failing to stop at an intersection, given the driver was on the cell
phone.
P(no stop|on cell) = 0.047/0.063 = 0.746
b) Compute the probability of failing to stop at an intersection, given the driver was not using a
cell phone.
P(no stop|not on cell) =0.201/.0.937 = 0.2145
c) Compute the probability of stopping properly at an intersection, given the driver was on the
cell phone.
P(stop|on cell) =0.016/0.063 = 0.254
d) Are cell phone usage and failing to stop at intersections independent of each other? Why or
why not?
They are not independent from each other since f(X,Y) ≠ f(X)f(Y)