Question

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?

Answer 1

(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"m" xx 2.0"m"//"pane"` and the rate of bubbles is `4 " 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" "~" POI"(lambda=2).`

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

`"Pr"(X>=1)" "=1-"Pr"(X=0)`

`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` `(i=1,2,3,4,5).` Then the `X_i stackrel "iid"" ~ ""POI"(2).` Assuming the glass panes are chosen independently,

`"Pr"(X_1=0, X_2=0, X_3=0, X_4=0, X_5=0)`

`=prod_(i=1)^5"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

`"Pr"(Y=y)" "=((n),(y))p^y(1-p)^(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 "iid"" ~ ""POI"(lambda), i=1,...,k`
then `Y=sum_(i=1)^kX_i" ~ ""POI"(klambda)`

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.