Can y'all help with this?
1 answer:
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Answer:
a. 0.1536
b. 0.9728
Step-by-step explanation:
The probability that a component fails, P(Y) = 0.2
The number of components in the system = 4
The number of components required for the subsystem to operate = 2
a. By binomial theorem, we have;
The probability that exactly 2 last longer than 1,000 hours, P(Y = 2) is given as follows;
P(Y = 2) = × 0.2² × 0.8² = 0.1536
The probability that exactly 2 last longer than 1,000 hours, P(Y = 2) = 0.1536
b. The probability that the system last longer than 1,000 hours, P(O) = The probability that no component fails + The probability that only one component fails + The probability that two component fails leaving two working
Therefore, we have;
P(O) = P(Y = 0) + P(Y = 1) + P(Y = 2)
P(Y = 0) = × 0.2⁰ × 0.8⁴ = 0.4096
P(Y = 1) = × 0.2¹ × 0.8³ = 0.4096
P(Y = 2) = × 0.2² × 0.8² = 0.1536
∴ P(O) = 0.4096 + 0.4096 + 0.1536 = 0.9728
The probability that the subsystem operates longer than 1,000 hours = 0.9728