Question

Ten percent of the engines manufactured on an assembly line are defective. What is the probability that the third nondefective engine will be found

(a) on the fifth trial?

(b) on or before the fifth trial?

(c) Given that the first two engines tested were defective, what is the probability that at least two more engines must be tested before the first nondefective is found?

(d) Find the mean and variance of the number of the trial on which a the first non-defective engine is found.

(e) Find the mean and variance of the number of the trial on which the third nonde-fective engine is found.

Answer 1

Solution

(a) on the fifth trial

Let X be the number of defective precede 3 nondefective .  \( X\sim Bb(3,0.9) \)

If 3 nondefective engines found in 5 trials \( \implies \)There are 2 defective 

\( \implies P(X=2)=C_4^2\left(0.9\right)^3\left(0.1\right)^2=0.6\left(0.9\right)^3\left(0.1\right)^2=0.043 \)

Therefore. \( P(X=2)=0.043 \)

(b) on or before the fifth trial

\( \implies P(X\leq 2)=(0.9)^3(1+0.3+6(0.1)^2) \)

                             \( =0.729(1.36) \)

                             \( =0.99144 \)

Therefore.  \( P(X\leq 2)=0.99144 \)

(c) Given that the first two engines tested were defective, what is the probability that

The objective is to find the probability that at least two more engines will be tested  

before the first non-defective is found given the first two engines tested are defective

 Let q be the probability of the engines manufactured on an assembly line are  defective = 0.10 

Let p be the probability of the engines manufactured on an assembly line are non defective = 0.90.

Let X represent the number of engines that are non-defective, and it follows a  geometric distribution with parameters p= 0.90 

The probability mass function of the geometric distribution is,

\( \implies P(X=x)=q^{x-1}P \hspace{2mm},x=1,2,3... \)

                             \( =\left(0.10\right)^{x-1}\left(0.9\right) \)

The cumulative distribution function is. 

\( \implies F(X)=P(X\leq x)=1-q^x=1-(0.10)^x \)

Use the lack of memoryless property of the geometric distribution to find the probability that at least two more engines will be tested before the first non-defective is found  

given the first two engines tested are defective.

The notational form of the required probability is. Take L.H.S

\( P(X\geq4)|X>2)=P(X>1) \)

\( \implies P(X\geq 4|x>2)=P(X>3|X>2) \)

                                        \( =\frac{P\left(X>3\cap X>2\right)}{P\left(X>2\right)}=\frac{P\left(X>3\right)}{P\left(X>2\right)} \)

                                        \( =\frac{1-P\left(X\le 3\right)}{1-P\left(X\le 2\right)}=\frac{1-\left[1-\left(0.10\right)^3\right]}{1-\left[1-\left(0.10\right)^2\right]}\hspace{5mm}\Big(F(X)=1-q^x\Big) \)

                                        \( =\frac{\left(0.10\right)^3}{\left(0.10\right)^2}=0.10=P\left(X>1\right)=R.H.S \)

Therefore.  \( P(X\geq 4|X>2)=0.10 \)

(d) Find the mean and variance of the number of the trial on which a the first non-defective engine is found.

Let the probability that non-defective engine found be  \( p=0.90\hspace{2mm},q=1-P=0.10 \)

we have  \( r=1 \)

\( Mean=\frac{r}{P}=\frac{1}{0.90}=1.1111 \)

\( Variance=\frac{rq}{P^2}=\frac{0.10}{(0.90)^2}=0.1235 \)

e) Find the mean and variance of the number of the trial on which the third nonde-fective engine is found.

Let the probability that non-defective engine found be  \( p=0.90\hspace{2mm},q=1-P=0.10 \)

we have  \( r=3 \)

\( Mean=\frac{r}{P}=\frac{3}{0.90}=3.3333 \)

\( Variance=\frac{rq}{P^2}=\frac{3\times 0.10}{(0.90)^2}=0.3704 \)

Therefore.

a). \( P(X=2)=0.043 \)

b). \( P(X\leq 2)=0.99144 \)

c). \( P(X\geq 4|X>2)=0.10 \)

d). \( Mean=1.1111 \),\( Variance=0.1235 \)

e). \( Mean=3.3333,Variance=0.3704 \)\( \)