HONR 399
September 13, 2010
Chapter 8 Answers
1. Consider the scenario in Example 8.1, in which we flip a coin three times. Let Y
denote the number of heads minus the number of tails. So, for instance, if (H, T, T ) is the
outcome, then Y = 1 − 2 = −1. Write a table that shows—for all of the 8 outcomes—the
values of Y . (It should have one column for the outcomes, and one column for Y , so your
table will look something like the table on page 4 of Chapter 8.)
outcome
(H, H, H)
(H, H, T )
(H, T, H)
(T, H, H)
(T, T, H)
(T, H, T )
(H, T, T )
(T, T, T )
Y
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1
1
1
−1
−1
−1
−3
2. Flip four fair coins, and let X denote the number of tails. Draw a picture of FX (x),
i.e., of the cumulative distribution function of X.
ru
llsl
I
+-a
llltl
?US
I t/t
Figure 1: The cdf for the number of tails in four fair coin flips.
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3. Consider example “28” on page 2 of Chapter 8, in which a randomly-chosen carton of
milk is selected. Let X be “1” if a carton of milk is 1% milkfat, or “2” if a carton of milk is
2% milkfat, or “3” if a carton of milk is fatfree, or “4” otherwise. Suppose that 13% of all
milk cartons have 1% milkfat, and 28% of all milk cartons have 2% milkfat, and 18% of all
milk cartons are fatfree, and 41% of all milk cartons are of some other type.
Compute the following:
P (X ≤ 1) = P (X = 1) = .13
P (X ≤ 2) = P (X = 1) + P (X = 2) = .13 + .28 = .41
P (X ≤ 3) = P (X = 1) + P (X = 2) + P (X = 3) = .13 + .28 + .18 = .59
P (X ≤ 4) = P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4) = .13 + .28 + .18 + .41 = 1
Draw a picture of FX (x), i.e., of the cumulative distribution function of X.
(,7
I
!,
b5'
GrrO
lh"
G-rCt
e-c
tl'
Figure 2: The cdf for the type of milk selected.
4. The Super Breakfast Challenge (SBC) consists of bacon, eggs, oatmeal, orange juice,
milk, and several other foods, and it costs $12.99 per person to order at a local restaurant.
It is known to be very difficult to consume the entire SBC. Only 10% of people are able to
eat all of the SBC. The other 90% of people will be unable to eat the whole SBC (it is too
much food!).
A probability student hears about the SBC and goes to the local restaurant. He observes
the number of customers, X, that attempt to eat the SBC, until the first success. So if there
are 4 failures and then 1 success (i.e., the outcome is (F, F, F, F, T )), then X = 5.
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(My wife says that this would be a perfect scenario for the show “Man Versus Food”!)
Compute the following:
Method #1
P (X ≤ 1) = .10
P (X ≤ 2) = .10 + (.90)(.10) = .19
P (X ≤ 3) = .10 + (.90)(.10) + (.90)2 (.10) = .271
P (X ≤ 4) = .10 + (.90)(.10) + (.90)2 (.10) + (.90)3 (.10) = .3439
P (X ≤ 5) = .10 + (.90)(.10) + (.90)2 (.10) + (.90)3 (.10) + (.90)4 (.10) = .40951
P (X ≤ 6) = .10+(.90)(.10)+(.90)2 (.10)+(.90)3 (.10)+(.90)4 (.10)+(.90)5 (.10) = .468559
P (X ≤ 7) = .10 + (.90)(.10) + (.90)2 (.10) + (.90)3 (.10) + (.90)4 (.10) + (.90)5 (.10) +
(.90)6 (.10) = .5217031
Method #2
P (X ≤ 1) = 1 − P (X > 1) = 1 − .90 = .10
P (X ≤ 2) = 1 − P (X > 2) = 1 − (.90)2 = .19
P (X ≤ 3) = 1 − P (X > 3) = 1 − (.90)3 = .271
P (X ≤ 4) = 1 − P (X > 4) = 1 − (.90)4 = .3439
P (X ≤ 5) = 1 − P (X > 5) = 1 − (.90)5 = .40951
P (X ≤ 6) = 1 − P (X > 6) = 1 − (.90)6 = .468559
P (X ≤ 7) = 1 − P (X > 7) = 1 − (.90)7 = .5217031
Draw a picture of FX (x), i.e., of the cumulative distribution function of X.
Figure 3: The cdf for the number of trials until the first SBC success.
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