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Difficulty: Very hardCompound Events and Probability Laws

Two independent security systems, XX and YY, operate in a facility. The probability that system YY detects an intrusion is 0.200.20 greater than the probability that system XX detects it. If the probability that at least one of the systems detects an intrusion is 0.920.92, what is the probability that exactly one of the systems detects the intrusion?

  1. 0.440.44Answer
  2. B
    0.480.48
  3. C
    0.520.52
  4. D
    0.320.32

Answer

The probability that exactly one of the systems detects the intrusion is 0.440.44.
The correct answer is 0.440.44. Setting P(X)=pP(X) = p and P(Y)=p+0.20P(Y) = p + 0.20, we use the independence rule P(X∩Y)=P(X)P(Y)P(X \cap Y) = P(X)P(Y) in the addition law P(X∪Y)=P(X)+P(Y)−P(X)P(Y)=0.92P(X \cup Y) = P(X) + P(Y) - P(X)P(Y) = 0.92. Solving p2−1.80p+0.72=0p^2 - 1.80p + 0.72 = 0 gives P(X)=0.60P(X) = 0.60 and P(Y)=0.80P(Y) = 0.80. The probability of both detecting the intrusion is 0.60×0.80=0.480.60 \times 0.80 = 0.48. Subtracting the probability of both from the probability of at least one (0.92−0.480.92 - 0.48) gives 0.440.44 for exactly one system detecting the intrusion.

Step-by-Step Solution

1
Define variables for the individual probabilities
Let P(X)=pP(X) = p. Then P(Y)=p+0.20P(Y) = p + 0.20.
System YY's probability is given as 0.200.20 greater than System XX's probability.
2
Apply the addition law for independent events
P(X∪Y)=P(X)+P(Y)−P(X∩Y)=p+(p+0.20)−p(p+0.20)=0.92P(X \cup Y) = P(X) + P(Y) - P(X \cap Y) = p + (p + 0.20) - p(p + 0.20) = 0.92
Since XX and YY are independent, P(X∩Y)=P(X)P(Y)P(X \cap Y) = P(X)P(Y).
3
Solve the quadratic equation for pp
2p+0.20−p2−0.20p=0.92  ⟹  −p2+1.80p−0.72=0  ⟹  p2−1.80p+0.72=02p + 0.20 - p^2 - 0.20p = 0.92 \implies -p^2 + 1.80p - 0.72 = 0 \implies p^2 - 1.80p + 0.72 = 0. Factoring yields (p−0.60)(p−1.20)=0(p - 0.60)(p - 1.20) = 0. Since p≤1p \le 1, p=0.60p = 0.60.
Probability values cannot exceed 11, so p=0.60p = 0.60 is the valid root.
4
Calculate individual probabilities and the probability of both occurring
P(X)=0.60P(X) = 0.60, P(Y)=0.80P(Y) = 0.80, and P(X∩Y)=0.60×0.80=0.48P(X \cap Y) = 0.60 \times 0.80 = 0.48.
These are needed to evaluate the compound probability of exactly one event occurring.
5
Calculate the probability that exactly one system detects the intrusion
P(exactly one)=P(X∪Y)−P(X∩Y)=0.92−0.48=0.44P(\text{exactly one}) = P(X \cup Y) - P(X \cap Y) = 0.92 - 0.48 = 0.44.
The probability of exactly one event occurring is the probability of at least one minus the probability of both.

Key Concept

Probability laws for independent and compound events
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Topics

Compound Events and Probability Laws

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