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Homework #3. Due: Friday, September 12, 2003.
<KEY> IE 230
Textbook: D.C. Montgomery and G.C. Runger, Applied Statistics and Probability for
Engineers, John Wiley & Sons, New York, 2003. Chapter 2, Sections 2.3–2.5.
1. Two useful probability results. Provide the reason for each step of the proofs.
(a) Result: P(A ) + P(A′ ) = 1.
Proof.
1
= P(S )
__ < Axiom 1 > __
= P(A ∪ A′ )
__ < Same event > __
= P(A ) + P(A′ )
__ < Axiom 3 > __
(b) Result: P(A ∪ B ) = P(A ) + P(B ) − P(A ∩B )
Proof.
P(A ∪ B )
= P[(A ∩B′ ) ∪ (A′ ∩ B ) ∪ (A ∩ B )]
__ < Same event > __
= P[(A ∩B′ ) ∪ (A′ ∩ B )] + P(A ∩ B )
__ < Axiom 3 > __
= P(A ∩B′ ) + P(A′ ∩ B ) + P(A ∩ B )
__ < Axiom 3 > __
= [P(A ∩B′ ) + P(A ∩ B )] + [P(A′ ∩ B ) + P(A ∩ B )] − P(A ∩ B )
__ < Algebra >__
= [P(A ∩B′ ) ∪ (A ∩ B )] + [P(A′ ∩ B ) ∪ (A ∩ B )] − P(A ∩ B )
__ < Axiom 3 > __
= P(A ) + P(B ) − P(A ∩ B )
__ < Same events > __
2. Consider two passengers with tickets for a particular air flight. Let Ai denote the event
that the i th passenger arrives to claim a seat. Suppose that P(A 1) = 0.9, P(A 2) = .95 and
P(A 1 ∩ A 2) = 0.88.
(a) In words, what is (A 1 ∩ A 2)?
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Both passengers arrive to claim a seat.
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(b) Determine the probability that at least one passenger arrives to claim a seat.
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P(A 1 ∪ A 2)
= P(A 1) + P(A 2) − P(A 1 ∩ A 1)
= 0.9 + 0.95 − 0.88
= 0.97 ←
Always true
Substitute given values
Simplify
(Comment: In the original version of this problem, P(A 1 ∩ A 2) = 0.8 was given,
leading to a probability greater than one.
Think about why that probability value is impossible.
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Homework #3. Due: Friday, September 12, 2003.
<KEY> IE 230
(c) Determine the probability that neither passenger arrives to claim a seat.
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P[(A 1 ∪ A 2)′]
= 1 − P(A 1 ∪ A 2)
= 1 − 0.97
= 0.03 ←
Complement
From Part (b)
Simplify
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(d) Draw a Venn diagram, showing in each of the four areas the corresponding
probability.
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Draw two overlapping circles in a rectangle. Label them A 1 and A 2.
In A 1 ∩ A 2, write 0.88.
In A 1 ∩ A 2′, write 0.02.
In A 1′ ∩ A 2, write 0.07.
In A 1′ ∩ A 2′ = (A 1 ∪ A 2)′, write 0.03.
(Comment: Can you argue carefully for each of these four values?)
(Comment: Can you redraw the Venn diagram so that the four areas are
proportional to the probabilities?)
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(e) Determine P(A 2 | A 1).
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= P(A 1 ∩ A 2) / P(A 1)
= 0.88 / 0.90
= 0.987 ←
P(A 2 | A 1)
Def. of conditional prob.
Substitute known values
Simplify
(Comment: Can you see the answer in the Venn Diagram of Part (d)?)
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(f) Determine P(A 2 | A 1′).
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P(A 2 | A 1′)
=
=
=
=
P(A 1′ ∩ A 2) / P(A 1′)
P(A 1′ ∩ A 2) / [1 − P(A 1)]
0.07 / (1 − 0.90)
0.7 ←
Def. of conditional prob.
Complement
Substitute known values
Simplify
(Comment: Can you see the answer in the Venn Diagram of Part (d)?)
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(g) Determine P(A 2′ | A 1).
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P(A 2′ | A 1)
= P(A 1 ∩ A 2′) / P(A 1)
= 0.02 / 0.90
= 0.222 ←
Def. of conditional prob.
Substitute known values
Simplify
(Comment: Can you see the answer in the Venn Diagram of Part (d)?)
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Homework #3. Due: Friday, September 12, 2003.
<KEY> IE 230
(h) Determine P(A 2′ | A 1′).
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P(A 2′ | A 1′)
=
=
=
=
=
P(A 1′ ∩ A 2′) / P(A 1′)
P[(A 1 ∪ A 2)′] / P(A 1′)
[1 − P(A 1 ∪ A 2)] / [1 − P(A 1)]
(1 − 0.97) / (1 − 0.90)
0.3 ←
Def. of conditional prob.
DeMorgan’s Law
Complements
Substitute known values
Simplify
(Comment: Can you see the answer in the Venn Diagram of Part (d)?)
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(i) Explain why P(A 2 | A 1) and P(A 2 | A′ 1) do not need to sum to one.
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Events are complementary only when discussed under the same conditioning event.
Here the conditions are complementary.
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3. (Montgomery and Runger, Problem 2–63) A lot contains 15 castings from a local supplier
and 25 castings from a supplier in the next state. Two castings are selected at random,
without replacement, from the lot of 40 castings. Let A be the event that the first casting
is selected from the local supplier and let B be the event that that second casting is
selected from the local supplier.
(a) What is the procedure that defines the experiment?
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Select one casting from the total lot of 40.
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(b) What sample space do you choose to use?
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In general, we could use the set of all pairs of castings.
(Comment: There are (40)(39) / 2 = 780 such pairs.
For Parts (c) and (d), however, it is simpler to use
smaller sample spaces specific to each part.)
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(c) Determine P(A ).
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Let the sample space be the set of forty castings.
Then because each casting is equally likely to be chosen,
P(A ) = 15 / 40 = 0.375 ←
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(d) Determine P(B | A ).
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Let the sample space be the set of the remaining 39 castings.
Then because each casting is equally likely to be chosen,
P(A ) = 14 / 39 = 0.359 ←
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Schmeiser
Homework #3. Due: Friday, September 12, 2003.
<KEY> IE 230
(e) Determine P(B ∩ A ).
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P(B ∩ A ) = P(A )P(B | A ) = (15 / 40)(14 / 39) = 0.135 ←
(Comment: We could have used the larger sample space
with 780 outcomes and counted the number of pairs for
which both are local. Evidently there are 105 such pairs.)
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(f) Determine P(B ∪ A ).
_____________________________________________________________
P(B ∪ A ) = P(B ) + P(A ) − (B ∩ A ) = (0.375) + (0.375) − 0.135 = 0.615 ←.
(Comment: Is it obvious that P(B ) = P(A )?
If not, use Total Probability to find
P(B ) = (B | A )P(A ) + (B | A′ )P(A′ )
= (14 / 39)(15 / 40) + (15 / 39)(25 / 40) = 0.375 = P(A ).)
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4. (Montgomery and Runger, Problem 2–73) Suppose that 2% of cotton fabric rolls and 3%
of nylon fabric rolls contain flaws. Of the rolls used by a manufacturer, 70% are cotton
and 30% are nylon.
(a) What procedure defines the experiment that underlies this problem?
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Randomly choose one of the fabric rolls.
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(b) Define notation for selecting a cotton roll, for selecting a nylon roll, and for selecting
a roll that contains flaws.
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Let C = "the roll is cotton."
Let F = "the roll contains flaws."
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(c) Write the given information in terms of your notation.
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P(C ) = 0.7 (and therefore P(C′ ) = 0.3)
P(F | C ) = 0.02
P(F | C′ ) = 0.03
(Comment: We could define N = "the roll is nylon", but C′ suffices.)
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(d) Determine the probability that a randomly selected roll contains flaws.
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P(F )
= P(F | C )P(C ) + P(F | C′ )P(C′ )
= (0.02)(0.7) + (0.03)(0.3)
= 0.023 ←
Total Probability
Substitute known values
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(e) Determine the probability that a randomly selected roll does not contain flaws.
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P(F ) = 1 − P(F ) = 1 − 0.023 = 0.097 ←
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Schmeiser
Homework #3. Due: Friday, September 12, 2003.
<KEY> IE 230
5. Consider the Monte Hall problem. Determine the probability of winning if the game has
four doors, there is only one prize, and Monte Hall opens two empty doors.
(a) If the contestant does not switch doors.
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As in class, let W = "the contestant wins" and
let C = "the contestant’s first choice is correct.
P(W )
= P(W | C )P(C ) + P(W | C′ )P(C′ )
= (1)(1 / 4) + (0)(3 / 4)
= 1/4←
Total Probability
Assume equally likely choices
simplify
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(b) If the contestant does switch doors.
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P(W )
= P(W | C )P(C ) + P(W | C′ )P(C′ )
= (0)(1 / 4) + (1)(3 / 4)
= 3/4←
Total Probability
Assume equally likely choices
simplify
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