- #1
carllacan
- 274
- 3
We have a sensor that measures a certain value that lies in the range (0, 3). The sensor is not perfect, and sometimes it fails. When this happens it ouputs a value under 1, regadless of the actual value. The failure probability is 0,01.
Suposing the sensor outputs a value under 1, what is the probability that the actual value is below 1? Suposing that we repeat the measurement N times and we always obtain a value under 1 what is the probability that the actual value is below 1?
My attempt at a solution:
Call a the actual value, and s the measured value. Then [itex]P(a<1) = \frac{1}{3}[/itex] and [itex]P(a>1) = \frac{2}{3}[/itex].Also [itex]P(s<1|a<1) = 1[/itex], [itex]P(s<1|a>1) = 0.01[/itex] and so [itex]P(s<1) = P(s<1|a<1)*P(a<1) + P(s<1|a>1)*P(a>1) = 1 \frac{1}{3}+0.01 \frac{2}{3} = 0.34[/itex] .
Using the Bayes Theorem [itex]P(a<1|s<1) = \frac{P(s<1|a<1)P(a<1)}{P(s<1)} = \frac{1/3}{0.34} = 0.98[/itex]
When we get to the N measuremetns case I start having problems. Intuitively I think that measurement failures are independent of each other, that is, a faulty first measurement does not influence the probability of a faulty second measurement, and so on. Then the answer would simply be [itex]P(a<1| s_{1:N} < 1) = 0.98^N[/itex]. It seems though misleadingly simple.
Could you help me out?
Suposing the sensor outputs a value under 1, what is the probability that the actual value is below 1? Suposing that we repeat the measurement N times and we always obtain a value under 1 what is the probability that the actual value is below 1?
My attempt at a solution:
Call a the actual value, and s the measured value. Then [itex]P(a<1) = \frac{1}{3}[/itex] and [itex]P(a>1) = \frac{2}{3}[/itex].Also [itex]P(s<1|a<1) = 1[/itex], [itex]P(s<1|a>1) = 0.01[/itex] and so [itex]P(s<1) = P(s<1|a<1)*P(a<1) + P(s<1|a>1)*P(a>1) = 1 \frac{1}{3}+0.01 \frac{2}{3} = 0.34[/itex] .
Using the Bayes Theorem [itex]P(a<1|s<1) = \frac{P(s<1|a<1)P(a<1)}{P(s<1)} = \frac{1/3}{0.34} = 0.98[/itex]
When we get to the N measuremetns case I start having problems. Intuitively I think that measurement failures are independent of each other, that is, a faulty first measurement does not influence the probability of a faulty second measurement, and so on. Then the answer would simply be [itex]P(a<1| s_{1:N} < 1) = 0.98^N[/itex]. It seems though misleadingly simple.
Could you help me out?
Last edited: