**Solutions to Homework: Section 2.3, pages 46-47,
#1, 3, 4, 8, 12**

**#1, page 46.** What is the probability that, on two
successive rolls of two dice, a sum of 7 comes up both times?

**SOLUTION:** We first need to compute the probability of rolling
a seven on a single roll of a pair of dice. We know that there are
36 possible outcomes of a roll of a pair of dice, and, from the table
on page 43 of the text, we see that exactly 6 of these outcomes
sum to seven. Thus, the probability of rolling a seven (on one roll
of a pair of dice) is 6 / 36 = 1 / 6, which is approximately
.167.

Since two rolls of a pair of dice are clearly independent, the probability of rolling a seven both times equals the probability of rolling a seven on the first roll times the probability of rolling a seven on the second roll. We just computed that each of these probabilities individually is 1 / 6 = .167, so the probability of rolling a seven on both rolls is (1 / 6) · (1 / 6) = 1 / 36, which is approximately (.167) · (.167) = .028.

**#3, page 47.** What is the probability that a
7 appears at least once in two rolls of a pair of dice? (Hint:
first find the probability that a 7 occurs in neither roll.)

**SOLUTION:** We start by addressing the hint. From
exercise #1 above, we know that the probability of rolling a seven on a
single roll of a pair of dice is 1 / 6, or approximately
.167. Therefore, the probability of *not* rolling a seven
on a single roll of a pair of dice is 1 – (1 / 6) = 5 / 6,
which is approximately 1 – .167 = .833. Following the same
reasoning as in exercise #1 above, since two rolls of a pair of dice are
clearly independent, the probability of *not* rolling a seven *either*
time equals the probability of *not* rolling a seven on the
first roll times the probability of *not* rolling a seven on the second
roll. That is, the probability of *not* rolling a seven
on *either* roll is (5 / 6) · (5 / 6) = 25 / 36,
which is approximately (.833) · (.833) = .694.

What does the hint have to do with the problem? We are trying
to compute the probability that we roll a seven *at least once* on
two rolls of a pair of dice; we could interpret this as a game in which
"winning" means rolling a seven at least once. Then not rolling a
seven at all would be losing the game, and we computed in the hint that
the probability of losing (not rolling a seven on either roll) is
25 / 36, or approximately .694. We know that the probability
of winning plus the probability of losing equals 1, from which we conclude
that the probability of winning (rolling a seven at least once) equals
1 – (probability of losing) = 1 – (25 / 36) = 11 / 36, which is approximately
1 – .694 = .306.

**#4, page 47.** Judy is learning to roll dice
Should she bet for or against a 7 appearing at least once in
three rolls of a pair of dice? Why? What about four rolls?

**SOLUTION:** This problem is a continuation of exercise
#3 above, in which we computed that the probability of rolling a seven
on *neither* of *two* rolls of a pair of dice is (5 / 6)
· (5 / 6) = 25 / 36, or approximately .694, and used
this to show that the probability of rolling a seven *at least once*
is 1 – (25 / 36) = 11 / 36, or approximately 1 – .694
= .306. How about on three rolls of a pair of dice? Since the
probability of *not* rolling a seven on a single roll is 5 /
6, as note above, and since the three rolls are certainly independent,
the probability of rolling a seven on *none* of these three rolls
is the product (5 / 6) · (5 / 6) · (5 / 6) = 125 /
216, which is approximately .579. Then, also as in exercise
#3 above, we see that the probability of rolling a seven at least once
in three rolls of the dice is 1 – (probability of not rolling a seven
at all) = 1 – 125 /216 = 91 / 216, which is approximately .421.
Thus, for three rolls, Judy is better off betting *against* rolling
a seven, since that has the greater probability.

Similarly, four rolls of a pair of dice are also clearly independent,
so the probability of rolling a seven on *none* of these four
rolls is the product (5 / 6) · (5 / 6) · (5 / 6) ·
(5 / 6) = 625 / 1296, which is approximately .482. Then,
as above, the probability of rolling a seven at least once in four rolls
of the dice is 1 – (probability of not rolling a seven at all) =
1 – 625 / 1296 = 671 / 1296, which is approximately .518. Thus,
for four rolls, Judy is better off betting *for* rolling a seven
at least once, since that has a slightly greater probability.

**#8, page 47.** A pair of dice is rolled three times.
What is the probability that neither 8, 9, nor
10 ever comes up? What is the probability that either
8, 9, or 10 comes up at least once?

**SOLUTION: **Following the ideas in the above exercises, we
start by computing the probabililty that an 8, 9, or
10 comes up on a single roll of a pair of dice. Using the table
on page 43 of the text, we count that there are five outcomes that sum
to 8, four that sum to 9, and three that sum to
10, so the probability of rolling an 8, 9, or
10 on a single roll of the dice is (5 + 4 + 3) / 36 = 12 /
36 = 1 / 3, which is approximately .333. Thus, the probability
of *not* rolling an 8, 9, or 10
on a single roll of the dice equals 1 – (probability of rolling an
8, 9, or 10) = 1 – (1 / 3) = 2 / 3, which is approximately
.667. Now we can use the fact that three rolls of a pair of dice
are certainly independent, to calculate that the probability of *not*
rolling an 8, 9, or 10 on three rolls of
the dice equals the product (2 / 3) · (2 / 3)
· (2 / 3) = 8 / 27, which is approximately .296.

Suppose now that we would like a sum of 8, 9, or
10 to come up *at least once* in three rolls of the dice.
If we think of this as a game in which we win if this happens, then losing
the game would mean that the sum never comes up 8, 9,
or 10 in the three rolls of the dice. But that is precisely
the probability that we computed in the first part of the exercise:
the probability of the dice *never* coming up 8, 9,
or 10 on three rolls of the dice is 8 / 27, or approximately
.296, and this is the probability of *losing* in this part of the
exercise. Therefore, the probability of winning (rolling 8,
9, or 10 at least once) equals 1 – (probability
of losing) = 1 – 8 / 27 = 19 / 27, which is approximately 1 –
.296 = .704.

**#12, page 47.** What is the probability that, in drawing
two cards from a standard deck, you get a queen and either the 5
of clubs or the 4 of hearts? (Here it does not matter
in which order you get the cards.)

**SOLUTION:** We cannot divide this game into independent tasks.
For example, if the first task is drawing one card from the standard deck,
and the second task is drawing another card from the standard deck, then
certainly the number of possibilities of winning the second task (drawing
a queen or the 5 of clubs or the 4 of hearts) depends
on whether or not one of these cards was drawn on the first draw.
So, we go back to the definition of probability. The total number
of possible outcomes is the number of possible two-card hands chosen from
a standard deck of 52 cards, which is _{52}C_{2}
= 52·51 / 2! = 1326. On the other hand, a winning hand consists
of one queen (chosen from the 4 queens in the deck) together
with either the 5 of clubs or the 4 of hearts.
Since order does not matter, in counting up the number of possible winning
hands we can choose the queen first (4 choices) and then choose
either the 5 of clubs or the 4 of hearts
(2 choices). Therefore, using the Fundamental Counting Principle,
we compute that there are 4·2 = 8 winning hands, which
means that the probability of winning is 8 / 1326, which is
approximately .0060.

**Solutions to Homework: Section 2.4, pages 52-54,
#1, 4, 5, 9, 10**

**#1, page 53.** Jack lures Ralph's suckers away by offering
them $200 if a 6 comes up and $600 if a
9 comes up. They still pay $100 each time to play.
What is Jack's expectation? Should Ralph try to lure them back by offering
$200 for a 6 and $650 for a 9?

**SOLUTION:** Jack receives $100 each time a player
rolls the dice, but what can Jack expect to have to pay the player? We
compute the *expected* (or average) payoff by multiplying each payoff by
its probability, and then adding. If the player rolls a six, the
payoff (for Jack) is

(5 / 36)
· (

That is, on the average Jack earns $5.56 per roll of the dice.

Can Ralph still expect a profit if he offers $200 for a 6
and $650 for a 9? The payoffs and probabilities are the
same for Ralph as they were for Jack, except when the player rolls a nine, in
which case the payoff for Ralph is

(5 / 36)
· (

That is, on the average Ralph would break even on this arrangement, so it is probably not worth his while to try.

**#4, page 53.** Big Jack has six $10 bills and
nine $5 bills all jumbled up in his pocket. He reaches in and
tips the waiter with one of the bills. What is his most probable tip?
What is his expected tip?

**SOLUTION:** Big Jack has 15 bills total in his pocket:
6 are $10 bills, and 9 are $5 bills.
So, the probability of choosing a $10 bill is *more probable* tip is $5. The *expected* (or
average) tip is computed by multiplying each payoff by its probability, and then
adding, so the expected tip is

(6 / 15)
· $10 + (9 / 15) · $5

So, on the average Big Jack will leave a tip of $7 (if he did this repeatedly).

**#5, page 53.** Elbert was one of 100,000
people to buy a raffle ticket for a $40,000 sports car. The
tickets cost 50¢ each. What is Elbert's most probable gain?
What is his expected gain? (We express a loss of $3 as a gain
of –$3.)

**SOLUTION:** Elbert spent $.50 for one ticket, and, if
he wins the raffle, he will receive a $40,000 car. So Elbert has two
possible payoffs: If he wins, the payoff is *more probable* gain is *expected* gain is computed by multiplying each payoff by its
probability, and then adding, so the expected gain is

(1 /
100,000) · $39,999.50 + (99,999 / 100,000) · (

So Elbert's expected "gain" is to lose 10¢.

**#9, page 54.** Diana tosses two coins with the following
payoffs. Compute her expectation in each case.

- Matching, 2¢, different, 1¢.

**SOLUTION:**There are 4 possible outcomes (namely,**HH**,**HT**,**TH**, and**TT**); 2 of them are matching (namely,**HH**and**TT**), and 2 of them are different (namely,**HT**and**TH**). Therefore, the probability of matching is2 / 4 = 1 / 2 , and the probability of different is also2 / 4 = 1 / 2 . So, Diana's*expectation*is(1 / 2) · 2¢ + (1/ 2) · 1¢ =(3 / 2)¢ = 1.5¢. That is, on the average Diana gets 1.5¢ per toss of the pair of coins. - At least one head, 5¢, no heads, 10¢.

**SOLUTION:**As in part (a), there are 4 possible outcomes; 3 of them contain at least one head (namely,**HH**,**HT**, and**TH**), and only 1 of them contains no head (namely,**TT**). Therefore, the probability of at least one head is3 / 4 , and the probability of no head is1 / 4 , so Diana's*expectation*is(3 / 4) · 5¢ + (1/ 4) · 10¢ = (25 / 4)¢ = 6.25¢. - At most one head, 2¢, two heads, 10¢.

**SOLUTION:**As in part (a), there are 4 possible outcomes; 3 of them contain at most one head (namely,**HT**,**TH**, and**TT**), and only 1 of them contains two heads (namely,**HH**). Therefore, the probability of at most one head is3 / 4 , and the probability of two heads is1 / 4 , so Diana's*expectation*is(3 / 4) · 2¢ + (1/ 4) · 10¢ = (16 / 4)¢ = 4¢.

**#10, page 54.** Elaine rolls one die and receives as many
dollars as the number that comes up. What is her expectation?

**SOLUTION:** There are 6 possible outcomes of a roll of
one die (namely, 1, 2, 3, 4, 5, and
6), and all are equally likely, so the probability of any one of them is
*expectation* is

(1 / 6) ·
$1

That is, on the average Elaine gets $3.50 per roll of the single
die.