This Exam P sample reference tests Conditional Probability. This is a conditional exchangeability problem. Once exactly three of six equally exposed units are known to be damaged, each three-unit subset is equally likely; four of the twenty subsets avoid both units in the distinguished pair, giving 0.20 and choice D.
How to solve this Conditional Probability question
Setup
Setup
Condition on exactly three damaged units among the six. Because every unit has the same independent damage probability, the identity of the damaged three is uniformly distributed over all three-unit subsets.
#{possible damaged subsets}=(36)=20
Model
Model
For neither member of the two-unit distinguished group to be damaged, all three damaged units must come from the other four.
#{favorable damaged subsets}=(34)=4
Compute
Compute
Take the ratio of favorable conditional subsets to all possible conditional subsets.
Pr(distinguished pair undamaged∣three damaged)
=(36)(34)=204=0.20
Answer
Answer
The required conditional probability is 0.2000.
0.2000(D)
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These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AThis is the unconditional joint probability that exactly three of the four nondistinguished units are damaged and every other unit is undamaged. It is the numerator before conditioning.
BThis result can be produced by retaining only one undamaged factor after selecting three damaged units from the group of four. It fails to account for both members of the distinguished pair.
CThis is the unconditional probability that exactly three of all six units are damaged. It is the denominator event, not the requested conditional ratio.
EThis is the unconditional probability that the two distinguished units are both undamaged. Knowing that three of the six are damaged changes that probability.
Original practice · fully worked
Original variant: navigation-beacon shock test
Eight sealed navigation beacons ride through the same shock test. Five use a standard casing and three use a ceramic casing. Each beacon independently fails with probability 0.20. After learning that exactly two beacons failed, calculate the probability that both failed beacons used the standard casing.
A 0.1049
B 0.2936
C 0.3571
D 0.3906
E 0.6250
Variant answer in brief
Given two failures among equally exposed beacons, every pair is equally likely. Ten of the 28 pairs come from the five standard-casing beacons, so the conditional probability is 5/14, approximately 0.3571, and choice C is correct.
Setup
Setup
Condition on the observed total of two failed beacons. Equal independent failure probabilities make the failed pair uniform among all pairs of the eight labeled beacons.
#{possible failed pairs}=(28)=28
Model
Model
A favorable pair contains two of the five standard-casing beacons.
#{favorable pairs}=(25)=10
Compute
Compute
The conditional probability is the favorable-pair count divided by the total pair count.
Pr(both standard∣two failures)=2810
2810=145=0.3571428571
Answer
Answer
Rounded to four decimal places, the result is 0.3571.
The 3108-page Probability Proof Manual reorganizes 718 verified Exam P solutions by syllabus skill and adds formula proofs, error patterns, and original worked practice.
