Worked solution · Exam P
This Exam P sample reference tests Inclusion–Exclusion. Adding the two union probabilities yields the identity 1+P(A), so P(A)=0.7+0.9-1=0.6, choice D.
Independent solution
Setup
Record the probabilities of the two unions that share event A and partition the remaining sample space by B and its complement.
Model
Adding the two union probabilities counts every outcome once and counts outcomes in A one additional time, giving 1+Pr(A).
Compute
Solving 0.7+0.9=1+Pr(A) gives Pr(A)=0.6.
Answer
The required marginal probability is 0.6, so the correct answer is choice D.
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Original practice · fully worked
A validation event A is split by whether a separate flag B is present. The recorded probabilities are P(A and B)=0.12 and P(A and not B)=0.28. Find P(A).
The two listed cells are disjoint and exhaust event A, so their sum is 0.40, choice D.
Setup
Partition event A into the disjoint cells where B occurs and where B does not occur.
Model
Because the two cells are disjoint and exhaust A, add their probabilities to recover the marginal probability of A.
Compute
Adding the two recorded cell probabilities gives 0.12+0.28=0.40.
Answer
Thus Pr(A)=0.40, which corresponds to choice D.
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