Original practice · fully worked
Original variant: failure of a four-module controller
A controller has four independent modules, each failing with probability 0.10. The controller shuts down if at least three modules fail. Find the shutdown probability.
- A
0.0010 - B
0.0028 - C
0.0037 - D
0.0100 - E
0.0523
Answer and calculation
The probabilities of exactly three and exactly four failures are 4(0.1³)(0.9)=0.0036 and 0.0001. Their sum is 0.0037, choice C.
Setup
Setup
Let the number of failed modules be binomial with four trials and failure probability 0.10.
K∼Binomial(4,0.10) Model
Model
Shutdown requires exactly three or exactly four failures.
Pr(K≥3)=(34)(0.1)3(0.9)+(0.1)4 Compute
Compute
The two probabilities are 0.0036 and 0.0001, giving a shutdown probability of 0.0037.
Pr(K≥3)=0.0036+0.0001=0.0037 Answer
Answer
The shutdown probability is 0.0037.
0.0037(C)