Original practice · fully worked
Original variant: locate a wildlife sighting by habitat zone
A reserve places 45% of its cameras in forest, 35% in marsh, and 20% on ridges. During one night, a camera in those zones records the target species with probabilities 0.10, 0.20, and 0.25. A randomly chosen camera produced a recording. Find the probability that it was a ridge camera.
- A
0.20000 - B
0.25000 - C
0.27273 - D
0.35000 - E
0.30303
Answer and calculation
Recording contributions are 0.045, 0.070, and 0.050. The ridge contribution is 10/33 of their total, choice E.
Setup
Setup
Multiply each habitat's camera share by its conditional recording probability to obtain three recording weights.
wF=0.45(0.10)=0.045 wM=0.35(0.20)=0.070 wR=0.20(0.25)=0.050 Model
Model
Condition on a recording by dividing the ridge weight by the sum of all three habitat weights.
Pr(R∣D)=wF+wM+wRwR Compute
Compute
The ridge contributes 0.050 of the total recording probability 0.165, so its posterior probability is 0.050/0.165=10/33.
Pr(R∣D)=0.050/0.165=10/33 Answer
Answer
The posterior probability of a ridge camera is 0.30303, which is choice E.
0.30303(E) 