Question

Safety and Reliability Engineering exercises Example 1 It has been known from experience that 4% of hard disk produced by a computer manufacturer are defective. Find a probability out of 50-disc testes what is the probability of having a a) 0 defects b) Are all defective p probabilitysuccess n independent trials 1-p failure P all are defective Q 0 defective I tried to find some formulas in google, but I didn't understand and didn't find anything. Binomial probability formula I don't know is this is right. Example 2 To ensure safe shut down of a nuclear power plant (NPP) during the normal and accidental conditions. There is a primary shut down system and a (missing word) secondary shut down system (SDS) is present (don't know if it's the right word) The failure probability of primarily SDS is 0.01 and secondary SDS is 0.035 Calculate the reliability of overall shut down system of nuclear power plant! Rs System Reliability Rs Ra Rb RaRb Rprim 1- 0.01-0.99 Rsecond 1-0.0350.965 Rs 0.990.965 (0.99x0.965) 0.99965 (I really don't know if it is even right!)

need help with safety and reliability exercises

Answer 1

Example 1.

Let's start from the beginning.

In case of Binomial Probability ,

Let X be a binomial Random Variable

where X denotes the number of defects obtained from the sample drawn

p = probability of success

q= 1-p= probability of failure.

n = number of independent trials.

The probability mass function this Binomial Random Variable X,

$P[X=x]=$$_{x}^{n}\textrm{C}.p^{x}.q^{n-x}$

Here,

n = 50

probability of being defective = 0.04 = p

probability of being non-defective= 1-0.04 =0.96=q

a)

The probability of having 0 defects,

$P[X=0]=$$_{x}^{n}\textrm{C}.p^{x}.q^{n-x}$

$=_{0}^{50}\textrm{C}. p^{0}.q^{50-0}$

  $=0.1298 =0.13$

b) The probability that all are defective,

$P[X=50]=$$_{x}^{n}\textrm{C}.p^{x}.q^{n-x}$

  $=_{50}^{50}\textrm{C}. p^{50}.q^{50-50}$

  $= (0.04)^{50} =1.268*10^{-70}$ [very small number]

Example 2.

To ensure safety shut down of Nuclear Power Plant ,  two system configurations are used.

1> Series System Configuration:-

Here,

failure probability of primary SDS=0.01

failure probability of secondary SDS= 0.035

$R_{prim}$$=1-.01=0.99$

$R_{second}= 1-.035=0.965$

Reliability of Overall Shut Down System of NPP=

$R_{Series}= \prod R_{prim}.R_{second}$

$=(0.99).(0.965) =0.95535$

$R_{S1}=0.95535$   

2.>

Parallel System Configuration:-

Here,

failure probability of primary SDS=0.01

failure probability of secondary SDS= 0.035

$R_{prim}$$=1-.01=0.99$

$R_{second}= 1-.035=0.965$

Reliability of Overall Shut Down System of NPP=

$R_{Parallel}= 1-\prod[1- R_{prim}].[1-R_{second}]$

$=1-[(1-0.99).(1-0.965)] =0.99965$

$R_{S2}=0.99965$