This Exam P sample reference tests Conditional Probability. Within the conditioning columns, the total probability is 0.86. The four cells with at least two disabilities contribute 0.12, so the conditional probability is 0.12/0.86=0.1395 and choice B.
How to solve this Conditional Probability question
Setup
Setup
Let A be the event of at least two disabilities and B the event of at most one death. The requested probability is a ratio within the columns defining B.
Pr(A∣B)=Pr(B)Pr(A∩B)
Model
Model
Sum every row in the zero- and one-death columns for the denominator, then restrict those columns to the two highest disability rows for the numerator.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AThe value 0.12 is the joint numerator P(A and B). It must still be divided by the conditioning probability 0.86.
CThe value 0.15 results from using an incomplete conditioning total of 0.80, since 0.12/0.80=0.15. Every table cell with zero or one death belongs in the denominator.
DThis interprets no more than one death as exactly one death: the qualifying numerator in that column is 0.05 and the column total is 0.20, giving 0.05/0.20=0.25.
EThe ratio 0.12/0.16=0.75 reverses the conditioning and computes the probability of at most one death given at least two disabilities.
Original practice · fully worked
Original variant: conditional alert attribution from a joint model
A monitoring system reports X critical faults, where X is 0, 1, or 2, and a rerouting flag Y, where Y is 0 or 1. Their joint pmf is p(x,y)=c(x+2y+1) on these six pairs. Given X+Y≥2, calculate P(Y=1).
A 1/4
B 1/3
C 1/2
D 2/3
E 3/4
Variant answer in brief
The conditioning event contains weights 3, 4, and 5 for pairs (2,0), (1,1), and (2,1). The two flagged pairs have total weight 9 of 12, so the conditional probability is 3/4 and choice E.
Setup
Setup
List the pairs satisfying the threshold and their unnormalized joint weights.
(2,0):3c,(1,1):4c,(2,1):5c
Model
Model
Within the conditioning event, the flag is present for the last two pairs. The common normalizing constant cancels.
Pr(Y=1∣X+Y≥2)=3c+4c+5c4c+5c
Compute
Compute
Sum the favorable and conditioning weights and reduce the ratio.
Pr(Y=1∣X+Y≥2)=129=43
Answer
Answer
The rerouting flag is present with conditional probability 3/4.
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