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 $10 m^{2}$ $10m^2$ . Consider glass pieces with dimensions 2.5m x 2.0m.?

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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 $10 m^{2}$ $10m^2$ . 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?

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Answer 1 · 2017-10-23T05:12:52

(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.

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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 $10 m^{2}$ $10m^2$ . Consider glass pieces with dimensions 2.5m x 2.0m.?

1 Answer

Explanation:

Alternate method for (ii):

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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^210m210m^2. Consider glass pieces with dimensions 2.5m x 2.0m.?

1 Answer

Explanation:

Alternate method for (ii):

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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 $10 m^{2}$ $10m^2$ . Consider glass pieces with dimensions 2.5m x 2.0m.?