Original practice · fully worked
Original variant: surviving components at a later checkpoint
At a maintenance check, three independent components with exponential lifetimes are all operating. For any one component, the chance of continuing for at least five more hours is 0.70. Ten hours later, what is the chance that the checkpoint shows precisely two components still operating?
- A
0.1176 - B
0.2401 - C
0.3674 - D
0.3823 - E
0.4900
Answer and calculation
Exponential survival over ten hours is the square of the stated five-hour survival, so each component remains with probability 0.49. A binomial calculation then gives 0.3674 for exactly two survivors and choice C.
Setup
Setup
Let p be the conditional probability that one currently operating component lasts another ten hours.
Pr(T>5)=e−5λ=0.70 Model
Model
The exponential survival function multiplies over two successive five-hour intervals.
p=e−10λ=(e−5λ)2=0.702=0.49 Compute
Compute
Independence makes the number of components still operating binomial with size three and success probability 0.49.
Pr(K=2)=(23)(0.49)2(0.51) =3(0.2401)(0.51)=0.367353 Answer
Answer
The probability of exactly two surviving components is about 0.3674.
0.3674(C) 