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Difficulty: MediumCompound Events and Probability Laws

During a quality control inspection at a manufacturing plant, two independent automated scanners, S1S_1 and S2S_2, inspect items along a production line. The probability that scanner S1S_1 detects a surface flaw on a defective item is 0.900.90, and the probability that scanner S2S_2 detects the flaw is 0.800.80. What is the probability that a surface flaw is detected by exactly one of the two scanners?

Answer: 0.26

Answer

The probability that a surface flaw is detected by exactly one of the two scanners is 0.26.
The probability of a surface flaw being detected by exactly one scanner is calculated by combining two mutually exclusive compound events: scanner S1S_1 detects it while scanner S2S_2 fails (0.90×0.20=0.180.90 \times 0.20 = 0.18), and scanner S1S_1 fails while scanner S2S_2 detects it (0.10×0.80=0.080.10 \times 0.80 = 0.08). Adding these two values yields 0.18+0.08=0.260.18 + 0.08 = 0.26.

Step-by-Step Solution

1
Determine the probabilities of non-detection for each scanner.
P(S_1') = 0.10, P(S_2') = 0.20
The probability of an event's complement is 1 minus the probability of the event.
2
Calculate the probability that scanner S_1 detects the flaw while scanner S_2 misses it.
P(S_1 \cap S_2') = 0.90 \times 0.20 = 0.18
Because the scanners operate independently, joint probability is the product of their individual probabilities.
3
Calculate the probability that scanner S_1 misses the flaw while scanner S_2 detects it.
P(S_1' \cap S_2) = 0.10 \times 0.80 = 0.08
Independence allows multiplication of individual probabilities.
4
Sum the probabilities of the two mutually exclusive outcomes.
P(\text{exactly one}) = 0.18 + 0.08 = 0.26
The addition law applies because detecting the flaw with only S_1 and only S_2 are mutually exclusive events.

Key Concept

Compound probability laws for independent and mutually exclusive events
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Topics

Compound Events and Probability Laws

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