1) A smoke detector uses two devices A and B. If smoke is present the probabilit
ID: 2923258 • Letter: 1
Question
1) A smoke detector uses two devices A and B. If smoke is present the probability that it will be detected by device A is 0.95, the probability it will be detected by device B is 0.90 and the probability it will be detected by both is 0.88
A) If smoke is present, and is detected by device A, find the probability of it being detected by device B
B) In addition to the proceeding information, consider the following facts:
- When smoke is not really present, there is a 0.02 probability of device A detecting smoke (a false alarm)
- The probability of smoke actually being present at any point in time is 0.005.
Given that device A has detected smoke, find the probability that smoke really is present
Explanation / Answer
P(A) = 0.95
P(B) = 0.9
P(A and B) = 0.88
(A)
Probability of detecting smoke by device B is independent of the probability of detecting smoke by device A.
Hence probability of detecting smoke by B given that A has detect the smoke is same as probability of detecting smoke by device B i.e. P(B) = 0.9
(B)
Probability of detecting smoke though it is not present, P(A|S') = 0.02
and P(A|S) = 0.98 (correct alarm produced)
P(S) = 0.005, P(S') = 0.995 (smoke not present)
P(S|A) = P(A|S)*P(S) / (P(A|S)*P(S) + P(A|S')*P(S'))
Required probability, P(S|A) = 0.98*0.005 / (0.99*0.005 + 0.02*0.995) = 0.1972
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