# 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?

(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

**(i)**

**(ii)**

**(iiI)**

#### Explanation:

**(i)**

Let

#"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

#"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" "~" 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

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.