# A process for making plate glass produces small bubbles (imperfections) scattered at random in the glass, at an average rate of 4 small bubbles per 10m^2. Consider glass pieces with dimensions 2.5m x 2.0m.?

## (i) Determine the probability that a piece of glass contains at least one small bubbles. (ii) Calculate the probability that 5 glass pieces chosen at random are all free of bubbles. (iii) Suppose 10 glass pieces were randomly chosen. What is the probability that 5 of the pieces contain at least 1 small bubble each?

Oct 23, 2017

(i) "Pr"(X>=1)=1-e^(–2)~~86.47%
(ii) prod_(i=1)^5"Pr"(X_i=0)=e^(–10)~~0.0045%
(iiI) "Pr"(Y=5)=252(1-e^(–2))^5e^(–10)~~0.553%

#### Explanation:

(i)

Let $X$ be the number of bubbles in a piece of glass. Then $X$ could be 0, 1, 2, etc. Since the glass is $2.5 \text{m" xx 2.0"m"//"pane}$ and the rate of bubbles is $4 {\text{ bubbles"//10"m}}^{2} ,$ we multiply these to get a rate parameter of lambda = 2" "("bubbles/pane"), meaning the model for this question is Poisson: $X \text{ "~" POI} \left(\lambda = 2\right) .$

"Pr"(X=x)" "=(e^(–lambda)lambda^x)/(x!)" "=" "(e^(–2)2^x)/(x!)

$\text{Pr"(X>=1)" "=1-"Pr} \left(X = 0\right)$

color(white)("Pr"(X>=1))" "=1-"(e^(–2)2^0)/(0!)

color(white)("Pr"(X>=1))" "=1-e^(–2)" "~~86.47%

(ii)

Let ${X}_{i}$ be the number of bubbles in glass piece $i$ $\left(i = 1 , 2 , 3 , 4 , 5\right) .$ Then the ${X}_{i} \stackrel{\text{iid"" ~ ""POI}}{2} .$ Assuming the glass panes are chosen independently,

$\text{Pr} \left({X}_{1} = 0 , {X}_{2} = 0 , {X}_{3} = 0 , {X}_{4} = 0 , {X}_{5} = 0\right)$

$= {\prod}_{i = 1}^{5} \text{Pr"(X_i=0)" }$(by independence)

=["Pr"(X_1=0)]^5" "(by identical distribution)

=[(e^(–2)2^0)/(0!)]^5

=e^(–10)" "~~0.0045%

(iii)

Let $Y$ be the number of panes of glass out of 10 that have at least one bubble. Since the probability of one pane containing at least one bubble was found to be 1-e^(–2), we can say that $Y$ has a Binomial distribution:

Y" "~" BIN"(n=10,"  "p=1-e^(–2))

and

$\text{Pr"(Y=y)" } = \left(\begin{matrix}n \\ y\end{matrix}\right) {p}^{y} {\left(1 - p\right)}^{n - y}$

color(white)("Pr"(Y=y))" "=((10),(y))(1-e^(–2))^y(e^(–2))^(10-y)

Then, the probability that exactly 5 pieces contain at least one bubble each is:

"Pr"(Y=5)" "=((10),(5))(1-e^(–2))^5(e^(–2))^(10-5)

color(white)("Pr"(Y=5))" "=252(1-e^(–2))^5e^(–10)

color(white)("Pr"(Y=5))" "~~0.553%

## Alternate method for (ii):

The sum of $k$ independent Poisson random variables (each with rate parameter $\lambda$) has a Poisson distribution itself. That is:

If ${X}_{i} \stackrel{\text{iid"" ~ ""POI}}{\lambda} , i = 1 , \ldots , k$
then $Y = {\sum}_{i = 1}^{k} {X}_{i} \text{ ~ ""POI} \left(k \lambda\right)$

So the chance of 5 pieces of glass containing a total of 0 bubbles across all of them is

"Pr"(Y=0)" "=" "(e^(–10)10^0)/(0!)

color(white)("Pr"(Y=0))" "=" "e^(–10)" "~~0.0045%

as before.