Math For Liberal Arts I

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 52C2 = 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  \$100 - \$200 = -\$100  (the  \$100  that Jack receives from the player, minus the  \$200  that Jack has to pay the player).  If the player rolls a nine, the payoff (for Jack) is  \$100 - \$600 = -\$500  (the  \$100  that Jack receives from the player, minus the  \$600  that Jack has to pay the player).  Otherwise, if the player rolls anything other than six or nine, the payoff (for Jack) is  \$100.  From the table on page 43, we see that the probability of rolling a six is  5 / 36  (since  5  of the  36  possible outcomes sum to six), the probability of rolling a nine is  4 / 36  (since  4  of the  36  possible outcomes sum to nine), and the probability of rolling something other than six or nine is  27 / 36  (since there are  27  remaining outcomes other than six or nine).  The expected payoff is then

(5 / 36) · (-\$100) + (4 / 36) · (-\$500) + (27 / 36) · \$100 = \$200 / 36 = \$5.56.

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  \$100 - \$650 = -\$550.  Therefore, the expected payoff for Ralph would be

(5 / 36) · (-\$100) + (4 / 36) · (-\$550) + (27 / 36) · \$100 = \$0.

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  6 / 15,  and the probability of choosing a  \$5  bill is  9 / 15.  Clearly the 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 = \$105 / 15 = \$7.

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 \$40,000 - \$.50 = \$39,999.50  (the value of the car minus the cost of the ticket),  while if he loses, the payoff is  –\$.50  (the cost of the ticket).  Since Elbert is one of  100,000  people who bought a ticket, his probability of winning is  1 / 100,000,  and hence his probability of losing is  99,999 / 100,000.  Clearly his more probable gain is  -\$.50.  The 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) · (-\$.50) = –\$10,000 / 100,000 = -\$.10.

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.

1. Matching,  2¢,  different,  1¢.
SOLUTION:  There are  4  possible outcomes (namely, HHHTTH,  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 is  2 / 4 = 1 / 2,  and the probability of different is also  2 / 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.
2. 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 is  3 / 4,  and the probability of no head is  1 / 4,  so Diana's expectation is  (3 / 4) · 5¢ + (1/ 4) · 10¢ = (25 / 4)¢ = 6.25¢.
3. 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 is  3 / 4, and the probability of two heads is  1 / 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  1 / 6.  Therefore, Elaine's expectation is

(1 / 6) · \$1 + (1 / 6) · \$2 + (1 / 6) · \$3 + (1 / 6) · \$4 + (1 / 6) · \$5 + (1 / 6) · \$6 = \$21 / 6 = \$3.50.

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