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?

1 Answer
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"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.