Question

In proof testing of circuit boards, the probability that any particular diode will fail is 0.01. Suppose a circuit board contains 200 diodes.

(a) How many diodes would you expect to fail, and what is the standard deviation of the number that are expected to fail?

(b) What is the (approximate) probability that at least four diodes will fail on a

(c) If five boards are shipped to a particular customer, how likely is it that at least four of them will work prop- erly? (A board works properly only if all its diodes work.)

 

 

Answer 1

Solution

we have  \( p(fial)=0.01 , 200 \) diodes  randomly selected board?

(a)  Find \( E(X)\hspace{2mm} and\hspace{2mm} \delta(X) \)

Let X : number of failing diodes. 

since.  \( X\sim Bin(200,0.01) \)

\( \implies E(X)=2 \hspace{3mm}and\hspace{3mm} V(X)=2\times 0.99=1.98 \)

\( \implies \delta(X)=\sqrt{V(X)}=\sqrt{1.98}=1.4 \)

Therefore.  \( E(X)=2\hspace{3mm},\delta(X)=1.4 \)

(b) probability that at least four diodes will fail

\( \implies P(X\geq 4)=\sum _{x=4}^{200}\:C_{200}^x\left(0.01\right)^x\left(0.99\right)^{n-x} \)

but since  \( n=200 \hspace{2mm}and \hspace{2mm} p\to 0 \hspace{2mm} and \hspace{2mm}np=1\times 2=2 \)

so we can approximate that the failed diodes is of rate \( 2, \lambda=2 \)

\( \implies X\sim P_0(\lambda)\implies P(X=x)=\frac{e^{-2}\times 2^x}{x!} \)

Therefore.  \( P(X\geq4)=1-\sum _{x=0}^3\:\frac{e^{-2}\times 2^x}{x!}=0.143 \)

(c) If five boards are shipped to a particular customer, how likely is it that at least four of them will work prop- erly? (A board works properly only if all its diodes work.)

All board work \( \implies P(w)=(0.99)^{200}=0.134 \)

Let X : number of diodes work. then  \( X\sim Bin(5,0.134) \)

\( \implies P(X\geq 4)=P(4)+P(5) \)

                             \( =4\times \left(0.134\right)^4\left(0.866\right)+\left(0.134\right)^5 \)

                             \( =0.001+0.00004=0.001 \)

Then  \( P(X\geq 4)=0.001 \)

Therefore. If is unlikely happen.

Therefore.

a). \( E(X)=2\hspace{3mm},\delta(X)=1.4 \)

b). \( P(X\geq4)=1-\sum _{x=0}^3\:\frac{e^{-2}\times 2^x}{x!}=0.143 \)

c). \( P(X\geq 4)=0.001 \)