Original practice · fully worked
Original variant: exclusive warnings from indicator covariance
During a diagnostic cycle, event A is a thermal warning and event B is a vibration warning. Their probabilities are 0.40 and 0.50. The covariance of the two event indicators is -0.02. Calculate the probability that exactly one warning occurs.
- A
0.18 - B
0.36 - C
0.50 - D
0.54 - E
0.72
Answer and calculation
Indicator covariance equals P(A and B)-P(A)P(B), so the overlap is -0.02+0.40(0.50)=0.18. Exactly one warning has probability P(A)+P(B)-2P(A and B)=0.54, which is choice D.
Setup
Setup
Let I_A and I_B be the two warning indicators and translate their product into the overlap event.
E[IA]=0.40,E[IB]=0.50 E[IAIB]=Pr(A∩B) Model
Model
Use the covariance to recover the otherwise unreported intersection.
−0.02=Pr(A∩B)−(0.40)(0.50) Pr(A∩B)=0.18 Compute
Compute
The sum of the marginals counts each exclusive warning once and the overlap twice, so remove the overlap twice.
Pr(exactly one)=Pr(A)+Pr(B)−2Pr(A∩B) =0.40+0.50−2(0.18)=0.54 Answer
Answer
Exactly one warning occurs with probability 0.54.
0.54(D)