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Math  /  Data & Statistics

QuestionQ2. An electrical system consists of four components as shown below. The system works if A and D and either B or C works. The probability of each component working is shown in the figure below.
Find the probability that the system works.

Studdy Solution

STEP 1

1. The probability of each component working is independent of the others.
2. The system works if components A and D work, and at least one of components B or C works.

STEP 2

1. Calculate the probability that both components A and D work.
2. Calculate the probability that at least one of components B or C works.
3. Combine these probabilities to find the probability that the system works.

STEP 3

Calculate the probability that both components A and D work.
The probability that A works is P(A)=0.8 P(A) = 0.8 .
The probability that D works is P(D)=0.7 P(D) = 0.7 .
Since A and D must both work, we multiply their probabilities:
P(A and D)=P(A)×P(D)=0.8×0.7=0.56 P(A \text{ and } D) = P(A) \times P(D) = 0.8 \times 0.7 = 0.56

STEP 4

Calculate the probability that at least one of components B or C works.
The probability that B works is P(B)=0.8 P(B) = 0.8 .
The probability that C works is P(C)=0.9 P(C) = 0.9 .
The probability that neither B nor C works is:
P(not B)=1−P(B)=0.2 P(\text{not } B) = 1 - P(B) = 0.2
P(not C)=1−P(C)=0.1 P(\text{not } C) = 1 - P(C) = 0.1
The probability that neither B nor C works is:
P(not B and not C)=P(not B)×P(not C)=0.2×0.1=0.02 P(\text{not } B \text{ and not } C) = P(\text{not } B) \times P(\text{not } C) = 0.2 \times 0.1 = 0.02
Therefore, the probability that at least one of B or C works is:
P(B or C)=1−P(not B and not C)=1−0.02=0.98 P(B \text{ or } C) = 1 - P(\text{not } B \text{ and not } C) = 1 - 0.02 = 0.98

STEP 5

Combine the probabilities to find the probability that the system works.
The system works if A and D work, and at least one of B or C works. Therefore, we multiply the probabilities:
P(System works)=P(A and D)×P(B or C)=0.56×0.98=0.5488 P(\text{System works}) = P(A \text{ and } D) \times P(B \text{ or } C) = 0.56 \times 0.98 = 0.5488
The probability that the system works is:
0.5488 \boxed{0.5488}

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